i want to share you about how to make graphic in coreldraw9.
The artistic media tool in CorelDRAW 9 was used to create the bubbles however, you could recreate the effect using clip art of bubbles, or by drawing your own bubbles.
1.) Draw a rectangle, this is the basis for our bar of soap.
2.) With the rectangle selected, move the pick tool over one of the corner nodes until the cursor changes (A), then drag the node inward to round the corners of the rectangle (B).
3.) Change the fill color to faded pink, and the line color to none. (A left click on the color palette changes the fill, a right click changes the line.)
5.) Click and drag on the rectangle with the extrude tool to set the vanishing point. You don't need to be precise here, we will make adjustments later.
6.) In the property toolbar, set the extrusion type for the bottom middle option. If this selection has a name, I'm not aware of it.
7.) Next, let's adjust the lighting so we can give our bar of soap some dimension. In the property bar select the lighting lighting control, and adjust lights 1 and 2 according to these settings:
8.) Adjust the bevel bevel control to a depth of 0.045" and an angle of 45 degrees.
9.) Now that it looks a bit more like a dimensional bar of soap, we can adjust the angle and depth of the extrusion to look a little more realistic. Click on the X in the middle of the bar of soap (the X indicates the vanishing point) and drag it into position so that your bar of soap looks about like mine.
Now for the fun part... bubbles! I have to admit I haven't really explored the artistic media sprayer in CorelDRAW very much yet. This is far from a scientific approach to doing this, but this is how I did it...
10.) Active the artistic media tool. In the property bar, choose the sprayer,
and activate the bubbles spray list.
11.) Draw a few random squiggley lines anywhere across the page and watch them turn into bubbles. We want our bubbles fairly dense, so be sure to draw a line with a lot of humps in it, like the one I've drawn.
12.) The bubbles are going to come out much larger than we need them, so just grab a corner handle and resize the group of bubbles. Keep drawing them, shrinking them, and putting them off to the side until you have a nice assortment of bubbles.
13.) Now activate the pick tool, right click on one set of bubbles and choose separate. You'll see the original line appear. Click on the line and delete it. Repeat this for all the other bubble groups.
14.) Next we are going to break these bubble groups up into even smaller bubble groups. Right click on each bubble group again, but this time chose ungroup (don't choose "ungroup all" or you will have a real mess on your hands!). After you ungroup each one, select smaller portions of the bigger group, and group those. We are just breaking the bubble groups up and putting them in smaller pieces. You can see from my example how I've broken them up.
15.) Now select all your bubble groups and make a copy of them once or twice so we have plenty of bubbles to work with. All these smaller groups of bubbles can be resized and rotated for variety.
16.) Begin to arrange them around your bar of soap so it looks like they are pooling around it. Remember to place a few bubbles behind the soap on the left and right to give it a sense of dimension. (To do this, right click on the bubble group, choose order -> to back). It's best to do this first.
17.) Then continue placing bubbles all around the bar of soap so it looks realistic. Don't forget you can duplicate, resize, and rotate the bubbles for more variety. Place a few tiny bubbles along the top edge of the bar of soap, too.
Here's how my soap suds looked after I finished placing them around the bar of soap.
If you have any extra bubbles leftover, you can just delete them. Now all we have left to do is add the text.
18.) We'll create the text off to the side, and then move it into place. Begin by clicking on the artistic text artisitic text tool tool, then click anywhere on the page, and type the text. Change the fill color to 70% black, center align it, adjust the size, and choose an appropriate font. You may need to use the shape tool to adjust the line and letter spacing.
19.) Select the text with the pick tool, then hit the + key on your keyboard to place a duplicate in the same position as the original. Use the arrow keys to nudge the duplicate down and to the right just a tiny bit. (You may need to go to options [ctrl-J] and adjust the nudge settings [under Workspace/Edit]. Mine is set to 0.01 inches.) Color the duplicate the same color as your bar of soap.
20.) Hit the + key again, color this duplicate white, nudge it down and to the right another tiny bit, and send it back one.
21.) Marquee select all three copies of the text, and move it into place on your bar of soap.
22.) Finally, export the image to the format of your choice. I didn't like the results exporting directly from CorelDRAW, so I opened the CDR file in Photo-Paint, and exported it from there.
Kamis, 15 Januari 2009
BAGAIMANA MANGUBAH SATUAN UKURAN
BAGAIMANA MANGUBAH SATUAN UKURAN
Rahasia Besar
Metode untuk mengubah unit-unit datang dari satu prinsip sederhana. Bilangan dengan satuan, seperti 16.2 meter atau 32 kaki/detik², tentu saja dapat dilakukan sama sebagai koefisien dengan variable, seperti 16.2x atau 32y/z².
Sekali kamu memahami ini, kamu melihat pada suatu kejadian mengapa hukum satuan bekerja seperti yang mereka lakukan. Kamu tidak dapat menambah 32 kaki ke 32 kaki/detik, kamu dapat menambah 32x ke 32x/y. dan ketika kamu membagi 32 mil oleh 4 jam untuk mendapatkan 8 mil/jam, hal itu tentu saja sama seperti membagi 32x oleh 4y sehingga diperoleh 8x/y.
Perkalian Oleh 1
Untuk mengubah satuan, hanya satu hal lain yang dibutuhkan untuk memuatnya dalam pikiran : kamu dapat mengalikan sesuatu dengan 1 dan itu tidak mengubah hasilnya. Jelas, benar? “jadi mengapa menyebutkannya?” aku mendengar kamu menjawab. Karena perkalian oleh 1 – bentuk pilihan yang hati-hati dari 1- adalah kunci untuk mengubah satuan.
Akan saya ilustrasikan dengan contoh. Secara bebas saya akan memilih yang mudah, satu yang mungkin dapat kamu lakukan dikepalamu, jadi saya dapat memperlihatkan langkah luar biasa dengan jelas.
Andaikan kamu ingin mengubah empat dan setengah jam ke menit. Tentu saja kamu tahu bahwa 60 menit = 1 jam. Sekarang bagilah keduanya dengan 1 jam. (ingat, kamu dapat melakukannya karena kamu menghilangkan satuan “jam” seperti variable. Jika kamu mempunyai 60x = 1y, pasti kamu dapat membagi kedua sisi dengan 1y.) setelah membagi, kamu punya 60 menit/1 = 1 jam. Mengapa saya melakukannya? Karena jika (60 menit)/(1 jam)=1, kemudian saya dapat mengalikan beberapa ukuran oleh pecahan dan tidak mengubah nilainya. (saya akan menjelaskan nanti mengapa saya membaginya terus dengan 1 jam dan bukan 60 menit.) jadi kembalilah kepada 4½ jam bahwa kita ingin mengubahnya ke dalam menit. Untuk melakukan konversi, perkalian sederhana oleh pilihan bagus yaitu 1 : 4.5 x 1 yang mana sama dengan 4.5 jam x (60 menit/1 jam). Sekarang, x kali y/z adalah sama dengan xy/z, jadi satuan kita sama dengan (4.5 jam x 60 menit)/1 jam. Perhatikan bahwa kamu mempunyai jam di keduanya, atas dan bawah. Kamu masih akan membaginya terus dengan x ketika x ada di kedua atas dan bawah, jadi kamu dapat membaginya terus dengan “variable” jam : (4.5 x 60 x menit) / 1 yang mana hasil perkaliannya 270 menit.
Ringkasan : untuk mengubah satuan, konseplah pecahan yang sama dengan 1, kalikan ukuran sebenarnya dengan pecahan dan sederhanakan.
Apakah Ini Benar-benar Perkalian oleh 1?
“Tapi tunggu sebentar!” saya mendengar kamu mengatakan. “Kamu memulai dengan 4.5 dan berakhir dengan 270. Bagaimana itu merupakan perkalian oleh 1?” jawabannya adalah bahwa kita tidak mulai dengan “dimensionless” angka murni 4.5, tetapi dengan 4.5 jam; dan kita tidak mengakhirinya dengan angka murni 270 tetapi 270 menit. Kamu harus bisa meyakinkan dirimu sendiri bahwa jika kamu membakar kalkun selama 270 menit atau 4½ jam, salah satu cara kamu menunggu di waktu yang sama ketika kamu makan malam.
Sebuah bilangan dengan satuan yang berbeda dari bilangan tanpa satuan atau dengan satuan yang berbeda, seperti 8x berbeda dari 8 dan 8y. pikirkan cara ini : 3.27 dollar atau euro sama dengan 327 sen, ketika kamu mengalikannya dengan “hati-hati memilih bentuk dari 1”, 100 sen/dollar atau 100 sen/euro. Jika yang atas dan bawah pecahan sama, pecahan sama dengan 1 dan nilai setelah kamu mengalikannya adalah sama dengan nilai sebelum dikalikan – tetapi di satuan yang berebeda, dimana tentu saja poin utuh.
Kamu harus berkata kepada dirimu sendiri,”Mengapa semua adalah pertentangan? Seseorang tahu bahwa untuk mengubah jam ke menit kamu harus mengalikannya dengan 60.”baik, iya, bahwa itu benar. Tetapi, dengan bebas pilih contoh sederhana untuk menunjukkan metodenya. Saya akan mencoba untuk lebih realistis contoh-contohnya dari sini.
Be Reasonable!
Jika kamu mengikuti prosedur pada halaman ini, akan tidak mungkin untuk membaginya dengan factor perubahan dimana kamu harus membagi atau “vice versa”.
Tetapi kita semua membuat kesalahan, jadi itu bagus untuk mempunyai keadaan kasar dan siap meneliti pekerjaanmu. Kamu dapat selalu menerapkan aturan ini : “jika tempatnya kecil, kamu membutuhkan lebih banyak dari itu untuk menjaga total yang sama.” Contohnya, setiap jam adalah lebih lama dari menit, jadi kamu mengharapkan 4.5 jam untuk mengubah beberapa angka yang lebih besar dari menit. Jika kamu mengakhirinya dengan 0.075 menit, kamu akan tahu kamu mempunyai kesalahan.
Di beberapa pekerjaan matematika, selalu terbaik untuk mengerjakan soal dengan dua cara yang berbeda, untuk berjaga-jaga terhadap ketidaktelitian. Tetapi kedua yang terbaik adalah mengerjakan soal dengan hati-hati dan menerapkan beberapa tes untuk kebijaksanaan, seperti yang ini.
Bagaimana cara memilih “1”
Kamu mungkin merasa heran bagaimana saya tahu untuk memilih bahwa pecahan tertentu yang sama dengan 1. Hanya ada dua langkah sederhana :
1. Temukan perubahan factor antara satuan yang diberikan dan satuan yang diminta, dan tulis itu sebagai persamaan.
Contoh : apakah kamu punya mill dan butuh kilometer, atau kamu punya kilometer dan butuh mill, kamu dapat menggunakan pengubahan factor yang lain antara mill dan kilometer, sebut saja 1 mill = 1.61 km atau 1 km = 0.621 mill. Persamaan yang lain akan bekerja lebih baik.
2. Ubah persamaan menjadi pecahan dengan satuan yang diminta pada pembilang dan satuan yang diberikan pada penyebut. Lebih formal, bagi kedua sisi dengan nilai dari sisi yang memuat satuan yang diberikan. (sebenarnya, aturan ini jauh lebih sederhana, seperti yang akan kita lihat di bawah.)
Contoh : untuk mengubah dari mil ke kilometer, kamu perlu pecahan dengan satuan yang diminta (kilometer) pada pembilangnya dan satuan yang diberikan (mil) pada penyebut. Berdasarkan pengubahan factor di atas, pecahan harus : 1.61 km/1 mil atau 1 km/ 0.621 mil.
Pecahan itu terlihat berbeda, tetapi ingatlah bahwa keduanya sama menuju 1 dan oleh karena itu mereka hanya berbeda bentuk dari pecahan yang sama. Yang lain akan bekerja hanya untuk menemukan perubahan.
Sekali kamu mempunyai bentuk pecahan fungsional tertentu dari 1, kalikan ukuran sebenarnya dengan pecahan dan sederhanakan. Contoh : jika ukuran sebenarnya adalah 15.7 mil, kamu akan mengalikannya dengan yang lain dari pecahan dan menghasilkan 25.3 km.
Dimana Dapat Menemukan Faktor-faktor Pengubah
Saya hanya menarik factor-faktor pengubah jauh dari topi saya. Banyak buku mempunyai table pengubah, termasuk Handbook of Chemistry and Physics. Ada juga beberapa sumber bagus. Favorit saya adalah US National Institute of Standards and Technology,http://ts.nist.gov/ts/htdocs/230/235/appxc.htm.
Carilah di beberapa referensi, kamu boleh mencatat artikel ini menggunakan singkatan biasa seperti sec (secon) dan hr (hours), daripada singkatan resmi (s dan h, berturut-turut). Hal itu disengaja, sejak kebanyakan siswa lebih mengenali bentuk yang lebih panjang. Dalam pekerjaan ilmiah, kamu diharapkan menggunakan bentuk yang resmi.
Rangkaian Peubah
Jika kamu dapat mengingat beberapa peubah, kamu mungkin bisa mengkombinasikannya untuk menghindari peubah spesifik. Jika kamu mempunyai kalkulator, itu dapat menjadi lebih cepat untuk melakukan ekstra aritmetika daripada mencari referensi dan mencari factor peubah tunggal. Contohnya, berapa meterkah 440 yard? Untuk mengubah 440 yard ke meter, kamu dapat mencari factor peubah antara yard dan meter. Tetapi jika kamu ingat bahwa 1in = 2.54 cm dan 36 in = 1 yd, itu mungkin lebih cepat untuk menggunakannya (ditambah 100 cm = 1 m) daripada mencari factor peubah tunggal. Ini berarti kamu kalikan dengan 3 perbedaan bentuk dari 1 : 440 yd x (36 in/1 yd) x (2.54 cm/1 in) x (1 m/100 cm) dan kelompokkan ke (440 x 36 x 2.54 yd in cm m)/(100 yd in cm). kerjakan aritmetika itu dan bagilah pembilang dan penyebutnya dengan yd,in, dan cm, kamu mempunyai jawabannya, 402 m. setelah memulainya dengan 440 yd dan mengalikannya dengan 1x1x1, kamu tahu bahwa nilai awalnya sama dengan nilai akhirnya : 440 yd = 402 m
Contohmu sendiri mungkin berbeda : kamu boleh mengingat peubah lain daripada yang saya kerjakan. Tetapi jika kamu melakukan sedikit factor untuk mengingatnya, kamu akan menemukan bahwa dengan mengkombinasikannya kamu dapat menghindari jumlah dari peubah-peubah.
Satuan Gabungan
Apakah satuan kompleks, seperti mengubah mill per jam ke kilometer per jam, atau bahkan mill per jam ke kaki (atau meter ) per detik? Kamu menggunakan teknik yang sama dan pengalikannya dengan pecahan tertentu yang sama dengan 1, hanya kamu yang perlu melakukannya untuk masing-masing satuan untuk dirubah. Itu hanya bentuk yang lebih umum dari rangkaiannya, dimana kamu siap mengetahui bagaimana melakukannya.
Contoh yang mengikutinya semakin membawamu kepada situasi yang lebih rumit :
1. Mil/jam km/jam menggambarkan peubah lurus dengan “per jam” satuan tidak berubah.
2. Mil/detik mil/jam
3. Km/jam m/detik menunjukkan bagaimana mengerjakan dua peubah pada kuantitas yang sama.
4. Sq ft sq m menunjukkan apa yang dikerjakan ketika satuan ditingkatkan pangkatnya.
Contoh 1 : mil per jam ke kilometer per jam
Soal ini dapat diselesaikan menggunakan 1 mil=1.61 km atau 1 km = 0.621 mil. Saya akan mengerjakannya dengan dua cara, dalam persamaan.
Untuk memulainya, tulis ukuran sebenarnya sebagai pecahan : 11.6 mil/jam
beranjak dari mil/jam ke km/jam, kamu melihat bahwa kamu mengakhirinya dengan penyebut yang sama saat kamu memulainya, jadi hanya pembilangnya yang mengubah satuan. Dengan kata lain, ini hanya teman lama kita milkilometer, dengan “per jam”
Rahasia Besar
Metode untuk mengubah unit-unit datang dari satu prinsip sederhana. Bilangan dengan satuan, seperti 16.2 meter atau 32 kaki/detik², tentu saja dapat dilakukan sama sebagai koefisien dengan variable, seperti 16.2x atau 32y/z².
Sekali kamu memahami ini, kamu melihat pada suatu kejadian mengapa hukum satuan bekerja seperti yang mereka lakukan. Kamu tidak dapat menambah 32 kaki ke 32 kaki/detik, kamu dapat menambah 32x ke 32x/y. dan ketika kamu membagi 32 mil oleh 4 jam untuk mendapatkan 8 mil/jam, hal itu tentu saja sama seperti membagi 32x oleh 4y sehingga diperoleh 8x/y.
Perkalian Oleh 1
Untuk mengubah satuan, hanya satu hal lain yang dibutuhkan untuk memuatnya dalam pikiran : kamu dapat mengalikan sesuatu dengan 1 dan itu tidak mengubah hasilnya. Jelas, benar? “jadi mengapa menyebutkannya?” aku mendengar kamu menjawab. Karena perkalian oleh 1 – bentuk pilihan yang hati-hati dari 1- adalah kunci untuk mengubah satuan.
Akan saya ilustrasikan dengan contoh. Secara bebas saya akan memilih yang mudah, satu yang mungkin dapat kamu lakukan dikepalamu, jadi saya dapat memperlihatkan langkah luar biasa dengan jelas.
Andaikan kamu ingin mengubah empat dan setengah jam ke menit. Tentu saja kamu tahu bahwa 60 menit = 1 jam. Sekarang bagilah keduanya dengan 1 jam. (ingat, kamu dapat melakukannya karena kamu menghilangkan satuan “jam” seperti variable. Jika kamu mempunyai 60x = 1y, pasti kamu dapat membagi kedua sisi dengan 1y.) setelah membagi, kamu punya 60 menit/1 = 1 jam. Mengapa saya melakukannya? Karena jika (60 menit)/(1 jam)=1, kemudian saya dapat mengalikan beberapa ukuran oleh pecahan dan tidak mengubah nilainya. (saya akan menjelaskan nanti mengapa saya membaginya terus dengan 1 jam dan bukan 60 menit.) jadi kembalilah kepada 4½ jam bahwa kita ingin mengubahnya ke dalam menit. Untuk melakukan konversi, perkalian sederhana oleh pilihan bagus yaitu 1 : 4.5 x 1 yang mana sama dengan 4.5 jam x (60 menit/1 jam). Sekarang, x kali y/z adalah sama dengan xy/z, jadi satuan kita sama dengan (4.5 jam x 60 menit)/1 jam. Perhatikan bahwa kamu mempunyai jam di keduanya, atas dan bawah. Kamu masih akan membaginya terus dengan x ketika x ada di kedua atas dan bawah, jadi kamu dapat membaginya terus dengan “variable” jam : (4.5 x 60 x menit) / 1 yang mana hasil perkaliannya 270 menit.
Ringkasan : untuk mengubah satuan, konseplah pecahan yang sama dengan 1, kalikan ukuran sebenarnya dengan pecahan dan sederhanakan.
Apakah Ini Benar-benar Perkalian oleh 1?
“Tapi tunggu sebentar!” saya mendengar kamu mengatakan. “Kamu memulai dengan 4.5 dan berakhir dengan 270. Bagaimana itu merupakan perkalian oleh 1?” jawabannya adalah bahwa kita tidak mulai dengan “dimensionless” angka murni 4.5, tetapi dengan 4.5 jam; dan kita tidak mengakhirinya dengan angka murni 270 tetapi 270 menit. Kamu harus bisa meyakinkan dirimu sendiri bahwa jika kamu membakar kalkun selama 270 menit atau 4½ jam, salah satu cara kamu menunggu di waktu yang sama ketika kamu makan malam.
Sebuah bilangan dengan satuan yang berbeda dari bilangan tanpa satuan atau dengan satuan yang berbeda, seperti 8x berbeda dari 8 dan 8y. pikirkan cara ini : 3.27 dollar atau euro sama dengan 327 sen, ketika kamu mengalikannya dengan “hati-hati memilih bentuk dari 1”, 100 sen/dollar atau 100 sen/euro. Jika yang atas dan bawah pecahan sama, pecahan sama dengan 1 dan nilai setelah kamu mengalikannya adalah sama dengan nilai sebelum dikalikan – tetapi di satuan yang berebeda, dimana tentu saja poin utuh.
Kamu harus berkata kepada dirimu sendiri,”Mengapa semua adalah pertentangan? Seseorang tahu bahwa untuk mengubah jam ke menit kamu harus mengalikannya dengan 60.”baik, iya, bahwa itu benar. Tetapi, dengan bebas pilih contoh sederhana untuk menunjukkan metodenya. Saya akan mencoba untuk lebih realistis contoh-contohnya dari sini.
Be Reasonable!
Jika kamu mengikuti prosedur pada halaman ini, akan tidak mungkin untuk membaginya dengan factor perubahan dimana kamu harus membagi atau “vice versa”.
Tetapi kita semua membuat kesalahan, jadi itu bagus untuk mempunyai keadaan kasar dan siap meneliti pekerjaanmu. Kamu dapat selalu menerapkan aturan ini : “jika tempatnya kecil, kamu membutuhkan lebih banyak dari itu untuk menjaga total yang sama.” Contohnya, setiap jam adalah lebih lama dari menit, jadi kamu mengharapkan 4.5 jam untuk mengubah beberapa angka yang lebih besar dari menit. Jika kamu mengakhirinya dengan 0.075 menit, kamu akan tahu kamu mempunyai kesalahan.
Di beberapa pekerjaan matematika, selalu terbaik untuk mengerjakan soal dengan dua cara yang berbeda, untuk berjaga-jaga terhadap ketidaktelitian. Tetapi kedua yang terbaik adalah mengerjakan soal dengan hati-hati dan menerapkan beberapa tes untuk kebijaksanaan, seperti yang ini.
Bagaimana cara memilih “1”
Kamu mungkin merasa heran bagaimana saya tahu untuk memilih bahwa pecahan tertentu yang sama dengan 1. Hanya ada dua langkah sederhana :
1. Temukan perubahan factor antara satuan yang diberikan dan satuan yang diminta, dan tulis itu sebagai persamaan.
Contoh : apakah kamu punya mill dan butuh kilometer, atau kamu punya kilometer dan butuh mill, kamu dapat menggunakan pengubahan factor yang lain antara mill dan kilometer, sebut saja 1 mill = 1.61 km atau 1 km = 0.621 mill. Persamaan yang lain akan bekerja lebih baik.
2. Ubah persamaan menjadi pecahan dengan satuan yang diminta pada pembilang dan satuan yang diberikan pada penyebut. Lebih formal, bagi kedua sisi dengan nilai dari sisi yang memuat satuan yang diberikan. (sebenarnya, aturan ini jauh lebih sederhana, seperti yang akan kita lihat di bawah.)
Contoh : untuk mengubah dari mil ke kilometer, kamu perlu pecahan dengan satuan yang diminta (kilometer) pada pembilangnya dan satuan yang diberikan (mil) pada penyebut. Berdasarkan pengubahan factor di atas, pecahan harus : 1.61 km/1 mil atau 1 km/ 0.621 mil.
Pecahan itu terlihat berbeda, tetapi ingatlah bahwa keduanya sama menuju 1 dan oleh karena itu mereka hanya berbeda bentuk dari pecahan yang sama. Yang lain akan bekerja hanya untuk menemukan perubahan.
Sekali kamu mempunyai bentuk pecahan fungsional tertentu dari 1, kalikan ukuran sebenarnya dengan pecahan dan sederhanakan. Contoh : jika ukuran sebenarnya adalah 15.7 mil, kamu akan mengalikannya dengan yang lain dari pecahan dan menghasilkan 25.3 km.
Dimana Dapat Menemukan Faktor-faktor Pengubah
Saya hanya menarik factor-faktor pengubah jauh dari topi saya. Banyak buku mempunyai table pengubah, termasuk Handbook of Chemistry and Physics. Ada juga beberapa sumber bagus. Favorit saya adalah US National Institute of Standards and Technology,http://ts.nist.gov/ts/htdocs/230/235/appxc.htm.
Carilah di beberapa referensi, kamu boleh mencatat artikel ini menggunakan singkatan biasa seperti sec (secon) dan hr (hours), daripada singkatan resmi (s dan h, berturut-turut). Hal itu disengaja, sejak kebanyakan siswa lebih mengenali bentuk yang lebih panjang. Dalam pekerjaan ilmiah, kamu diharapkan menggunakan bentuk yang resmi.
Rangkaian Peubah
Jika kamu dapat mengingat beberapa peubah, kamu mungkin bisa mengkombinasikannya untuk menghindari peubah spesifik. Jika kamu mempunyai kalkulator, itu dapat menjadi lebih cepat untuk melakukan ekstra aritmetika daripada mencari referensi dan mencari factor peubah tunggal. Contohnya, berapa meterkah 440 yard? Untuk mengubah 440 yard ke meter, kamu dapat mencari factor peubah antara yard dan meter. Tetapi jika kamu ingat bahwa 1in = 2.54 cm dan 36 in = 1 yd, itu mungkin lebih cepat untuk menggunakannya (ditambah 100 cm = 1 m) daripada mencari factor peubah tunggal. Ini berarti kamu kalikan dengan 3 perbedaan bentuk dari 1 : 440 yd x (36 in/1 yd) x (2.54 cm/1 in) x (1 m/100 cm) dan kelompokkan ke (440 x 36 x 2.54 yd in cm m)/(100 yd in cm). kerjakan aritmetika itu dan bagilah pembilang dan penyebutnya dengan yd,in, dan cm, kamu mempunyai jawabannya, 402 m. setelah memulainya dengan 440 yd dan mengalikannya dengan 1x1x1, kamu tahu bahwa nilai awalnya sama dengan nilai akhirnya : 440 yd = 402 m
Contohmu sendiri mungkin berbeda : kamu boleh mengingat peubah lain daripada yang saya kerjakan. Tetapi jika kamu melakukan sedikit factor untuk mengingatnya, kamu akan menemukan bahwa dengan mengkombinasikannya kamu dapat menghindari jumlah dari peubah-peubah.
Satuan Gabungan
Apakah satuan kompleks, seperti mengubah mill per jam ke kilometer per jam, atau bahkan mill per jam ke kaki (atau meter ) per detik? Kamu menggunakan teknik yang sama dan pengalikannya dengan pecahan tertentu yang sama dengan 1, hanya kamu yang perlu melakukannya untuk masing-masing satuan untuk dirubah. Itu hanya bentuk yang lebih umum dari rangkaiannya, dimana kamu siap mengetahui bagaimana melakukannya.
Contoh yang mengikutinya semakin membawamu kepada situasi yang lebih rumit :
1. Mil/jam km/jam menggambarkan peubah lurus dengan “per jam” satuan tidak berubah.
2. Mil/detik mil/jam
3. Km/jam m/detik menunjukkan bagaimana mengerjakan dua peubah pada kuantitas yang sama.
4. Sq ft sq m menunjukkan apa yang dikerjakan ketika satuan ditingkatkan pangkatnya.
Contoh 1 : mil per jam ke kilometer per jam
Soal ini dapat diselesaikan menggunakan 1 mil=1.61 km atau 1 km = 0.621 mil. Saya akan mengerjakannya dengan dua cara, dalam persamaan.
Untuk memulainya, tulis ukuran sebenarnya sebagai pecahan : 11.6 mil/jam
beranjak dari mil/jam ke km/jam, kamu melihat bahwa kamu mengakhirinya dengan penyebut yang sama saat kamu memulainya, jadi hanya pembilangnya yang mengubah satuan. Dengan kata lain, ini hanya teman lama kita milkilometer, dengan “per jam”
A SQUARE
A SQUARE
The definition of a square is a rectangle that a pair of the close each others sides are congruent. For example, the description of the picture, if ABCD is a square, then = = = and all of every single angles are .
A square has two diagonals, they are and , where is diagonal = diagonal . If we cut the square at one point of the diagonal, we will be obtained two right equilateral triangles. So, we can find the side of the square if we know the length of its diagonal.
Example :
AC = 10, find the sides of square!
Answer
The side of the square are = = = = x
ADC is a right angle.
To find the side of square we use phytagoras theorem because ADC ia a right angle.
Look the picture!
Then
So, the sides of the square are 5
Exercise
Find the side of the square if =
The definition of a square is a rectangle that a pair of the close each others sides are congruent. For example, the description of the picture, if ABCD is a square, then = = = and all of every single angles are .
A square has two diagonals, they are and , where is diagonal = diagonal . If we cut the square at one point of the diagonal, we will be obtained two right equilateral triangles. So, we can find the side of the square if we know the length of its diagonal.
Example :
AC = 10, find the sides of square!
Answer
The side of the square are = = = = x
ADC is a right angle.
To find the side of square we use phytagoras theorem because ADC ia a right angle.
Look the picture!
Then
So, the sides of the square are 5
Exercise
Find the side of the square if =
VIDEO
VIDEO _
13. The figure above shows the graph of y equal g for x. if the function h is defined by h for x equal g for two x plus two. What is the value of h for one?
Problem Solving :
- We are looking for h for one and the first information of the graph listing like it ______
- The next information is
h(x) = g(2x) + 2
- Then substitute 1 to this equation
h(1) = g(2) + 2
- If g(2) = y where is 2 = x, so we can find the value of y by graph. When x = 2 , so we get y = 1. Then we substitute it in the equation
h(1) = 1 + 2
= 3
13. Let the function f be defined by f for x equal x plus one. If two times f for p equals twenty, what is the value of f for three p?
Problem Solving :
- We are looking for f for three p. in another word, “What is f when x equal three p?”
- First information is f for x equal x plus one and the next information is two times f for p equal twenty.
- We start with equation here.
2 f(p) = 20
- Then the function divide by two
F(p) = 10
- What is the p for the situation f for x equal x plus one
f(p) = p + 1 = 10
p = 9
- It is no the answer, because we looking for f when x equal three p
- If we take p equal nine and substitute to x equal three p, so x equal twenty seven. And we have question for function f which is f for x equal x plus one and now we know if
x = 27, so f(27) = 27 + 1
f(27) = 28 (this is the answer)
17. In the xy-coordinate plane, the graph of x = y2 – 4 intersect line l at (0,p) and (s,t). what is the greatest possible value of the slope of l?
Problem Solving :
- We will looking for the greatest slope m.
We can see it peace by peace.
In the xy-coordinate plane the graph of x = y2 + 4 and it is the graph ________
- Whether 2 value that formed by x and x intersect in four.
- Intersect line l in intersection at two points (o,p) and (5,t).
- We write two points that important
______________
And the asked what is the greatest possible value of the slope of l.
- What do we know about the slope will we know that the slope of the line is going to be
Line l = m =
- We applied (0,p) and (5,t) to the line
m =
VIDEO _
FACTORING POLYNOMIALS
- One way defined factors of the polynomial is true of form algebraic long division that look suspiciously like a long division you learned the kids, only harder.
- For example, lets try the shift x – 3 a factor of x3 – 7x – 6
- When dividing x – 3 in to x3 – 7x – 6
First set up the problem make a long division problem for elementary school that is you dividing x – 3 into x3 – 7x – 6
Zero in there is no second degrees term.
- Now you must ask your self what times x give you x cube of course x square. So you write x2 as the polemic question and then multiply x – 3 by x2, which give you x3 – 3x2 which is subtract from x3 + 0x2 to get 3x2.
- Bring it down the next term, make it -7x you have 3x2 – 7x. Now we begin again dividing x – 3 into 3x2 – 7x. Just looking at the first terms, x skill into 3x2 3x times.
- Now 3x is the next part of the answer multiply x – 3 by 3x for applied at all 3x2 – 9x. Subtracting you get 2x – 6. Now we see that x – 3 devise evenly into 2x – 6 which is close 2 without remainder. So the solution to the long division problem x3 - 7x – 6 divide by x – 3 is x2 + 3x + 2.
- Since evenly without reminder then x – 3 is a factor of x3 – 7x – 6
- x2 + 3x + 2 is also a factor of x3 – 7x – 6
- we now know that
x3 – 7x – 6 = (x - 3)( x2 + 3x + 2)
- The quadratic expression (x2 + 3x + 2) can be factored into (x + 1)(x + 2), so
x3 – 7x – 6 = (x - 3) (x + 1)(x + 2)
- Setting the factored formed the equation x3 – 7x – 6 to zero. We get
0 = (x - 3) (x + 1)(x + 2)
Those ever x – 3 = 0
X + 1 = 0
X + 2 = 0
Solving all of this equation for x we get x = 3, x = -1, or x = -2
The roots x3 – 7x – 6 is 3, -1, -2
- Now the three roots for this 3rd degree equation. You remember that the quadratic (2rd degree) equations always have at most 2 roots
- A 4th degree equation would have 4 or fewer roots
- The degree of polynomial equation always limits the number of roots.
- Let summarized long division process for a 3rd order polynomial
1. Find a partial quotient of x2 by dividing x into x3 to get x2
2. Multiply x2 by the divisor and subtract the product from the dividend.
3. Repeat the process until you either “clear it out” or reach a reminder.
VIDEO _
PRE CALCULUS
- Graph of rational function which can have discontinuities has a polynomial in the denominator. Which mean you are dividing by something that is very quantity? So you can be sure what the bottom of the fraction is.
- It is possible that some value of x will meet to division by zero. Example :
f(x) =
and when x = 1 the function value becomes
f(x) =
which is f(x) = , with zero in the denominator.
For this function choosing x = 1 is the bad idea.
- What the bad choice when we make the bottom of the rational function zero. It shows a break in function graph. For example :
Suppose f(x) =
Insert 0 for x, f(x) = = = -2
So, you can put it down on the graph.
- Next to try x = 1, you get f(1) = = , it is impossible.
That means, you can not compute the way value when x = 1.
- Rational functions don’t always work this way.
- Take the graph f(x) =
For this case, what number you choose for x, the denominator never can to be zero. All rational functions will give zero in denominator.
- A break can show up in two ways. The simply type of break is missing point on the graph. For example : y =
- The graph loses like these if x – 3. If you try to substitute x – 3 to equation,
y = = , not allowed.
That is not possible, not feasible, and not allowed.
So, there is no way for x – 3. This is typical example of the missing point syndrome.
- When you see result of and also tell you direction is possible to factor top and bottom of the rational function and simplify.
- Example : y = , the top factor to (x - 3)(x + 2)
y =
the x – 3 on the top cancels with the bottom. So, how function simplifies to
y = x + 2
- Missing point is a loophole. The rational function without simplify, x – 3 is bad point because it means that division by zero. But, if we simplify first, it is no problem for x – 3. That is y = x + 2
13. The figure above shows the graph of y equal g for x. if the function h is defined by h for x equal g for two x plus two. What is the value of h for one?
Problem Solving :
- We are looking for h for one and the first information of the graph listing like it ______
- The next information is
h(x) = g(2x) + 2
- Then substitute 1 to this equation
h(1) = g(2) + 2
- If g(2) = y where is 2 = x, so we can find the value of y by graph. When x = 2 , so we get y = 1. Then we substitute it in the equation
h(1) = 1 + 2
= 3
13. Let the function f be defined by f for x equal x plus one. If two times f for p equals twenty, what is the value of f for three p?
Problem Solving :
- We are looking for f for three p. in another word, “What is f when x equal three p?”
- First information is f for x equal x plus one and the next information is two times f for p equal twenty.
- We start with equation here.
2 f(p) = 20
- Then the function divide by two
F(p) = 10
- What is the p for the situation f for x equal x plus one
f(p) = p + 1 = 10
p = 9
- It is no the answer, because we looking for f when x equal three p
- If we take p equal nine and substitute to x equal three p, so x equal twenty seven. And we have question for function f which is f for x equal x plus one and now we know if
x = 27, so f(27) = 27 + 1
f(27) = 28 (this is the answer)
17. In the xy-coordinate plane, the graph of x = y2 – 4 intersect line l at (0,p) and (s,t). what is the greatest possible value of the slope of l?
Problem Solving :
- We will looking for the greatest slope m.
We can see it peace by peace.
In the xy-coordinate plane the graph of x = y2 + 4 and it is the graph ________
- Whether 2 value that formed by x and x intersect in four.
- Intersect line l in intersection at two points (o,p) and (5,t).
- We write two points that important
______________
And the asked what is the greatest possible value of the slope of l.
- What do we know about the slope will we know that the slope of the line is going to be
Line l = m =
- We applied (0,p) and (5,t) to the line
m =
VIDEO _
FACTORING POLYNOMIALS
- One way defined factors of the polynomial is true of form algebraic long division that look suspiciously like a long division you learned the kids, only harder.
- For example, lets try the shift x – 3 a factor of x3 – 7x – 6
- When dividing x – 3 in to x3 – 7x – 6
First set up the problem make a long division problem for elementary school that is you dividing x – 3 into x3 – 7x – 6
Zero in there is no second degrees term.
- Now you must ask your self what times x give you x cube of course x square. So you write x2 as the polemic question and then multiply x – 3 by x2, which give you x3 – 3x2 which is subtract from x3 + 0x2 to get 3x2.
- Bring it down the next term, make it -7x you have 3x2 – 7x. Now we begin again dividing x – 3 into 3x2 – 7x. Just looking at the first terms, x skill into 3x2 3x times.
- Now 3x is the next part of the answer multiply x – 3 by 3x for applied at all 3x2 – 9x. Subtracting you get 2x – 6. Now we see that x – 3 devise evenly into 2x – 6 which is close 2 without remainder. So the solution to the long division problem x3 - 7x – 6 divide by x – 3 is x2 + 3x + 2.
- Since evenly without reminder then x – 3 is a factor of x3 – 7x – 6
- x2 + 3x + 2 is also a factor of x3 – 7x – 6
- we now know that
x3 – 7x – 6 = (x - 3)( x2 + 3x + 2)
- The quadratic expression (x2 + 3x + 2) can be factored into (x + 1)(x + 2), so
x3 – 7x – 6 = (x - 3) (x + 1)(x + 2)
- Setting the factored formed the equation x3 – 7x – 6 to zero. We get
0 = (x - 3) (x + 1)(x + 2)
Those ever x – 3 = 0
X + 1 = 0
X + 2 = 0
Solving all of this equation for x we get x = 3, x = -1, or x = -2
The roots x3 – 7x – 6 is 3, -1, -2
- Now the three roots for this 3rd degree equation. You remember that the quadratic (2rd degree) equations always have at most 2 roots
- A 4th degree equation would have 4 or fewer roots
- The degree of polynomial equation always limits the number of roots.
- Let summarized long division process for a 3rd order polynomial
1. Find a partial quotient of x2 by dividing x into x3 to get x2
2. Multiply x2 by the divisor and subtract the product from the dividend.
3. Repeat the process until you either “clear it out” or reach a reminder.
VIDEO _
PRE CALCULUS
- Graph of rational function which can have discontinuities has a polynomial in the denominator. Which mean you are dividing by something that is very quantity? So you can be sure what the bottom of the fraction is.
- It is possible that some value of x will meet to division by zero. Example :
f(x) =
and when x = 1 the function value becomes
f(x) =
which is f(x) = , with zero in the denominator.
For this function choosing x = 1 is the bad idea.
- What the bad choice when we make the bottom of the rational function zero. It shows a break in function graph. For example :
Suppose f(x) =
Insert 0 for x, f(x) = = = -2
So, you can put it down on the graph.
- Next to try x = 1, you get f(1) = = , it is impossible.
That means, you can not compute the way value when x = 1.
- Rational functions don’t always work this way.
- Take the graph f(x) =
For this case, what number you choose for x, the denominator never can to be zero. All rational functions will give zero in denominator.
- A break can show up in two ways. The simply type of break is missing point on the graph. For example : y =
- The graph loses like these if x – 3. If you try to substitute x – 3 to equation,
y = = , not allowed.
That is not possible, not feasible, and not allowed.
So, there is no way for x – 3. This is typical example of the missing point syndrome.
- When you see result of and also tell you direction is possible to factor top and bottom of the rational function and simplify.
- Example : y = , the top factor to (x - 3)(x + 2)
y =
the x – 3 on the top cancels with the bottom. So, how function simplifies to
y = x + 2
- Missing point is a loophole. The rational function without simplify, x – 3 is bad point because it means that division by zero. But, if we simplify first, it is no problem for x – 3. That is y = x + 2
Five lose myth on the subject of mathematics
INEQUALITY
A. Definition
1. Inequality Notation
Assume if a and b are real number.
a. a is called less than b, is written a < b if and only if a - b is negative. For example, 7 < 12 because 7 – 12 = -5 and -5 is negative.
b. A is called more than b, is written a > b if and only if a – b is positive. For example, 5 > 2 because 5 – 2 = 3, and 3 is positive.
c. A is called less than or equals by b, is written a ≤ b if and only if b if and only if a < b or a = b. In another word, a ≤ is negation from a > b. for example, 4 ≤ 7 is true because 4 > 7 is false.
d. A is called more than or equals by b, is written a ≥ b if only if a > b or a = b. in another word a ≥ b is negation from a < b. for example, 7 ≥ 3 is true because 7 < 3 is false.
2. Inequality Definition
Above description, were given notation from … a ≤ b and a ≥ b. inequality is an open sentence that related by notation “<”, “>“, “≤“, or “≥”. The general forms of inequality are
u (x) < v (x) u (x) ≤ v (x)
u (x) > v (x) u (x) ≥ v (x)
Example 4.1
An inequality form’s example
a. X – 2 < 5 d. l5 – 3xl < 13
b. x² - 1 ≥ 2 e.
c. (x – 5)(x + 1)²(x + 3) > 0
3. Interval
Before we learn about solution of inequality form, do we learn about solution of inequality form, do you still remember about an interval? For example R is a real number set. We can explain a set that is collection of the real number x… For example, {x l x < 4, x Є R}, {x l x ≥ 1, x Є R}, {x l 2 < x ≤ 5, x Є R}. the sets can be explain as bellow.
There are 8 kind of interval forms possibility that often we find if solve an inequality generally.
Table 4.1
Set Notation Number Line Kind of Interval
1. {x l x < a, x Є R}
2. {x l x > a, x Є R}
3. {x l x ≤ a, x Є R}
4. {x l x ≥ a, x Є R}
5. {x l a < x < b, x Є R}
6. {x l a < x ≤ b, x Є R}
7. {x l a ≤ x < b, x Є R}
8. {x l a < x < b, x Є R} Opened interval at a
Opened interval at a
Closed interval at b
Closed interval at b
Opened interval at a and b
Opened interval at a and closed interval at b
Closed interval at a and opened interval at b
Closed interval at a and b
Interval in 1-4 is named infinite interval, whereas interval in 5-8 is named finite interval.
4. Inequality Characteristic
Interval definition explain presence a solve from the inequality. Determine of the solve or the solution is based of theorems bellow.
- Theorem 4.1
Interval notation an inequality is not change if there is added or reduction a number (statement) at both of parts of an equation that same.
- Proof
Let c is a number (statement) which will added in a < b inequality. Because of a < b, so a – b = n, n < 0. Thus
a – b + c – c = n, c – c = 0
⇔ (a + c) – (b + c)=n < 0
⇔ a + c < b + c
Example 4.2
Determine the solution of each inequality bellow!
a. X – 2 < 5
b. X – 4 > -6
c. X + 9 < 11
Answer :
a. X – 2 < 5 b. x – 4 > -6 c. x + 9 < 11
⇔ x – 2 + 2 < 5 + 2 ⇔ x – 4 – 4 > -6 + 4 ⇔ x + 9 – 9 < 11 – 9
⇔ x < 7 ⇔ x > 2 ⇔ x < 2
Note :
Let x – 2 < 5
The left part of inequality above is x – 2, whereas the right part of inequality above is 5.
- Theorem 4. 2
Notation an inequality is not change if make a positive number (statement) multiplication or division in both of parts of an inequality.
- Proof
Let a < b and k > 0. Because of a < b, so a – b = n, n < 0
Thus
k(a - b) = kn
⇔ ka – kb = kn < 0
⇔ ka < kb
Example 4.3
Determine the solution of each inequality bellow!
a. 2x > 14
b. 3x < 18
c. 2x + 3 < 9
Answer
a. 2x > 14 b. 3x < 18 c. 2x + 3 < 9
⇔ ½ . 2x > ½ . 4 ⇔ ⇔ 2x + 3 – 3 < 9 - 3
⇔ x > 7 ⇔ x < 6 ⇔ 2x < 6
⇔
⇔ x < 3
- Theorem 4.3
Notation of inequality is change if make a negative number (statement) multiplication or division in both of parts of an inequality.
- Proof
For example a < b and k < 0
Because of a < b, so a – b = n, n < 0
Thus
k(a - b) = kn
⇔ ka – kb = kn > 0
⇔ ka > kb
Example 4.4
a. -5x < 10
b. -4x > -20
c. -3x + 5 < x – 1
Answer
a. -5x < 10 b. -4x > -20 c. -3x + 5 < x - 1
⇔ ⇔ ⇔ -3x + 5 – 5 < x – 1 – 5
⇔ x > -2 ⇔ x < 5 ⇔ -3x < x – 6
⇔ -3x – x < x –x – 6
⇔ -4x <-6
⇔
⇔ x > 3/2
Playing a number
Is 7 < 4 true ?
1. Let a = 3 and b = 4, thus : a < b
2. Multiplying by a : a² < ab
3. Reducing by b² : a² - b² < ab - b²
4. Make a factor : (a +b)(a - b) < b (a - b)
5. Over by (a – b) : a + b < b
6. Replacing a with 3 and b with 4 : 7 < 4
Where are the false of argument above?
B. Linear Inequality
1. Definition
Linear inequality is inequality that contain variable with its highest power is 1 (one). Example 4.2 – 4.4 showed same linear inequality forms with variable x.
2. Solve the Linear Inequality
In this topic, our term is only for one variable linear inequality. The solution linear inequality is divided become two, they are :
a. Absolute linear inequality
This inequality is true to all of its variables. For example, x + 2 < x + 5
b. Conditional Linear inequality
This inequality is true to certain value of its variable. For example, x – 2 > 8 is true for x > 10
Solution of this linear inequality can showed by a set or a number line (interval) like was showed in example 4.2 – 4.4.
Example 4.5
Determine the solution set in number line bellow of inequalities bellow!
a. 4x + 2 < 10
b. 7x – 5 ≥ x – 17
c. 3x + 5 < 5x + 7
Answer
a. 4x + 2 < 10 c. 3x + 5 < 5x – 7
⇔ 4x + 2 – 2 < 10 – 2 ⇔ 3x + 5 – 3x - 7 < 5x + 7 – 3x - 7
⇔ 4x < 8 ⇔ -2 < 2x
⇔ ⇔ -1 < x
⇔ x < 2 SS = {x l x > -1, x Є R}
SS = {x l x < 2, x Є R}
b. 7x – 5 ≥ x – 17
⇔ 7x – 5 + 5 ≥ x – 17 + 5
⇔ 7x ≥ x – 12
⇔ 7x – x ≥ x – x – 12
⇔ 6x ≥ -12
⇔
⇔x ≥ -2
SS = {x l x ≥ -2, x Є R}
Exercise 4.1
In exercise 1-8, observant
1. x < 6 ⇔ x + 2 < 8 5. a > 3 ⇔ 5a > 15
2. y > 8 ⇔ y – 5 > -2 6. P < -4 ⇔ 3p > 3p > -12
3. z – 1 ≤ 4 ⇔ x ≤ 5 7. n ≥ 2 ⇔ -3n ≥ -6
4. m + 2 ≥ 1 ⇔ m ≥ -1 8. m > 7 ⇔ -m > -7
In exercise 9-20, determine inequality set to all of real number component by set notation!
9. 5 ≤ k 13. 8 > q + 6 17.
10. a < -2 14. 5n < 20 18. -3 < m + 2 < 4
11. p + 1 < 3 15. –w ≥ 20 19. 2x – 1 ≤ 3x + 5 ≤ 2x + 6
12. t – 2 > 1 16. 20. x – 1 ≤ 2x + 1 ≤ 3 + x
In exercise 21-30, determine the set solution to all of real number component in a number line.
21. x < 1 25. 10 – 2z ≥ 6 28. 2x – 1 < x
22. 2y + 1 < 3 26. 5p – 6 > 4 29. 4(y - 2) ≤ 0
23. 2 + 5k ≤ 12 27. 2m + 1 < 0 30. 3 < 2x – 7 ≤ 5
24. 1 – p > 2 + 7p
A. Definition
1. Inequality Notation
Assume if a and b are real number.
a. a is called less than b, is written a < b if and only if a - b is negative. For example, 7 < 12 because 7 – 12 = -5 and -5 is negative.
b. A is called more than b, is written a > b if and only if a – b is positive. For example, 5 > 2 because 5 – 2 = 3, and 3 is positive.
c. A is called less than or equals by b, is written a ≤ b if and only if b if and only if a < b or a = b. In another word, a ≤ is negation from a > b. for example, 4 ≤ 7 is true because 4 > 7 is false.
d. A is called more than or equals by b, is written a ≥ b if only if a > b or a = b. in another word a ≥ b is negation from a < b. for example, 7 ≥ 3 is true because 7 < 3 is false.
2. Inequality Definition
Above description, were given notation from … a ≤ b and a ≥ b. inequality is an open sentence that related by notation “<”, “>“, “≤“, or “≥”. The general forms of inequality are
u (x) < v (x) u (x) ≤ v (x)
u (x) > v (x) u (x) ≥ v (x)
Example 4.1
An inequality form’s example
a. X – 2 < 5 d. l5 – 3xl < 13
b. x² - 1 ≥ 2 e.
c. (x – 5)(x + 1)²(x + 3) > 0
3. Interval
Before we learn about solution of inequality form, do we learn about solution of inequality form, do you still remember about an interval? For example R is a real number set. We can explain a set that is collection of the real number x… For example, {x l x < 4, x Є R}, {x l x ≥ 1, x Є R}, {x l 2 < x ≤ 5, x Є R}. the sets can be explain as bellow.
There are 8 kind of interval forms possibility that often we find if solve an inequality generally.
Table 4.1
Set Notation Number Line Kind of Interval
1. {x l x < a, x Є R}
2. {x l x > a, x Є R}
3. {x l x ≤ a, x Є R}
4. {x l x ≥ a, x Є R}
5. {x l a < x < b, x Є R}
6. {x l a < x ≤ b, x Є R}
7. {x l a ≤ x < b, x Є R}
8. {x l a < x < b, x Є R} Opened interval at a
Opened interval at a
Closed interval at b
Closed interval at b
Opened interval at a and b
Opened interval at a and closed interval at b
Closed interval at a and opened interval at b
Closed interval at a and b
Interval in 1-4 is named infinite interval, whereas interval in 5-8 is named finite interval.
4. Inequality Characteristic
Interval definition explain presence a solve from the inequality. Determine of the solve or the solution is based of theorems bellow.
- Theorem 4.1
Interval notation an inequality is not change if there is added or reduction a number (statement) at both of parts of an equation that same.
- Proof
Let c is a number (statement) which will added in a < b inequality. Because of a < b, so a – b = n, n < 0. Thus
a – b + c – c = n, c – c = 0
⇔ (a + c) – (b + c)=n < 0
⇔ a + c < b + c
Example 4.2
Determine the solution of each inequality bellow!
a. X – 2 < 5
b. X – 4 > -6
c. X + 9 < 11
Answer :
a. X – 2 < 5 b. x – 4 > -6 c. x + 9 < 11
⇔ x – 2 + 2 < 5 + 2 ⇔ x – 4 – 4 > -6 + 4 ⇔ x + 9 – 9 < 11 – 9
⇔ x < 7 ⇔ x > 2 ⇔ x < 2
Note :
Let x – 2 < 5
The left part of inequality above is x – 2, whereas the right part of inequality above is 5.
- Theorem 4. 2
Notation an inequality is not change if make a positive number (statement) multiplication or division in both of parts of an inequality.
- Proof
Let a < b and k > 0. Because of a < b, so a – b = n, n < 0
Thus
k(a - b) = kn
⇔ ka – kb = kn < 0
⇔ ka < kb
Example 4.3
Determine the solution of each inequality bellow!
a. 2x > 14
b. 3x < 18
c. 2x + 3 < 9
Answer
a. 2x > 14 b. 3x < 18 c. 2x + 3 < 9
⇔ ½ . 2x > ½ . 4 ⇔ ⇔ 2x + 3 – 3 < 9 - 3
⇔ x > 7 ⇔ x < 6 ⇔ 2x < 6
⇔
⇔ x < 3
- Theorem 4.3
Notation of inequality is change if make a negative number (statement) multiplication or division in both of parts of an inequality.
- Proof
For example a < b and k < 0
Because of a < b, so a – b = n, n < 0
Thus
k(a - b) = kn
⇔ ka – kb = kn > 0
⇔ ka > kb
Example 4.4
a. -5x < 10
b. -4x > -20
c. -3x + 5 < x – 1
Answer
a. -5x < 10 b. -4x > -20 c. -3x + 5 < x - 1
⇔ ⇔ ⇔ -3x + 5 – 5 < x – 1 – 5
⇔ x > -2 ⇔ x < 5 ⇔ -3x < x – 6
⇔ -3x – x < x –x – 6
⇔ -4x <-6
⇔
⇔ x > 3/2
Playing a number
Is 7 < 4 true ?
1. Let a = 3 and b = 4, thus : a < b
2. Multiplying by a : a² < ab
3. Reducing by b² : a² - b² < ab - b²
4. Make a factor : (a +b)(a - b) < b (a - b)
5. Over by (a – b) : a + b < b
6. Replacing a with 3 and b with 4 : 7 < 4
Where are the false of argument above?
B. Linear Inequality
1. Definition
Linear inequality is inequality that contain variable with its highest power is 1 (one). Example 4.2 – 4.4 showed same linear inequality forms with variable x.
2. Solve the Linear Inequality
In this topic, our term is only for one variable linear inequality. The solution linear inequality is divided become two, they are :
a. Absolute linear inequality
This inequality is true to all of its variables. For example, x + 2 < x + 5
b. Conditional Linear inequality
This inequality is true to certain value of its variable. For example, x – 2 > 8 is true for x > 10
Solution of this linear inequality can showed by a set or a number line (interval) like was showed in example 4.2 – 4.4.
Example 4.5
Determine the solution set in number line bellow of inequalities bellow!
a. 4x + 2 < 10
b. 7x – 5 ≥ x – 17
c. 3x + 5 < 5x + 7
Answer
a. 4x + 2 < 10 c. 3x + 5 < 5x – 7
⇔ 4x + 2 – 2 < 10 – 2 ⇔ 3x + 5 – 3x - 7 < 5x + 7 – 3x - 7
⇔ 4x < 8 ⇔ -2 < 2x
⇔ ⇔ -1 < x
⇔ x < 2 SS = {x l x > -1, x Є R}
SS = {x l x < 2, x Є R}
b. 7x – 5 ≥ x – 17
⇔ 7x – 5 + 5 ≥ x – 17 + 5
⇔ 7x ≥ x – 12
⇔ 7x – x ≥ x – x – 12
⇔ 6x ≥ -12
⇔
⇔x ≥ -2
SS = {x l x ≥ -2, x Є R}
Exercise 4.1
In exercise 1-8, observant
1. x < 6 ⇔ x + 2 < 8 5. a > 3 ⇔ 5a > 15
2. y > 8 ⇔ y – 5 > -2 6. P < -4 ⇔ 3p > 3p > -12
3. z – 1 ≤ 4 ⇔ x ≤ 5 7. n ≥ 2 ⇔ -3n ≥ -6
4. m + 2 ≥ 1 ⇔ m ≥ -1 8. m > 7 ⇔ -m > -7
In exercise 9-20, determine inequality set to all of real number component by set notation!
9. 5 ≤ k 13. 8 > q + 6 17.
10. a < -2 14. 5n < 20 18. -3 < m + 2 < 4
11. p + 1 < 3 15. –w ≥ 20 19. 2x – 1 ≤ 3x + 5 ≤ 2x + 6
12. t – 2 > 1 16. 20. x – 1 ≤ 2x + 1 ≤ 3 + x
In exercise 21-30, determine the set solution to all of real number component in a number line.
21. x < 1 25. 10 – 2z ≥ 6 28. 2x – 1 < x
22. 2y + 1 < 3 26. 5p – 6 > 4 29. 4(y - 2) ≤ 0
23. 2 + 5k ≤ 12 27. 2m + 1 < 0 30. 3 < 2x – 7 ≤ 5
24. 1 – p > 2 + 7p
Minggu, 21 Desember 2008
TUGAS I
A lot of lose myths about mathematics. This lose myths give great contribution in make partly of people allergy dislike mathematics. Finally, majority of our students got bad score in this subject, not because of disable, but because of early they feel allergy and be afraid, so they never or lazy to study this subject. Although there are many lose myths about mathematics, but there are five lose myths that have grew on and create bad perception to mathematics.
The first myth, mathematics is science that difficult until there are a little people who capable to comprehend by minimal IQ certain. It is mislead of course. Although it is not the easiest science, actually mathematics is a science that more easy than other science. For example, observe this problem comparison for the students of private elementary school grade six. The first problem is “Mention 3 Central Kalimantan traditional dances.” Second problem is “A circle is divided become 3 sections with comparison of center angle are 2:3:4, then count the value of center angle of the sections.”
Be in fact, the percentage of students who have right answer in second problem more than the percentage of students who have right answer in first problem. Without cause controvertion to occur, the samples above indicate if mathematics is not difficult science. The students regard as mathematics problems is difficult because they not comprehend numeral concept and measure concept correctly when they were elementary school students. If the numeral concept and the measure concept is governable, so analyzing and counting become easy and enjoyable.
The second myth, mathematics is memorized science from the many much formula. This myth made students lazy to study mathematics and finally not understand about mathematics. Whereas, substantively the mathematics is not the formula’s memorized science, because without comprehend the concept, the formulas that have memorized not useful. For example, there is a problem, “The machine is assembled by Benny longer 6 hour than Ahmad. If they can assemble the machine together in 4 hour, how long time needed by Ahmad to assemble the machine?”
Someone who memories quadrate equation will not able to answer those problem if they not capable take it to quadrate equation’s type. Actually, only a little mathematics’s formula needed to remind, whereas much of the other formulas not necessary remind, but it is comprehend enough. The example, if students understand the anatomy concept form cone slice, so more than 90% the cone slice’s formulas not remind.
The third myth, mathematics is always related to speed of calculate. The calculating belong mathematics, especially in the elementary school. But, capability of calculate quickly is not the most important in mathematics. The most important is concept comprehend. By concept comprehend, we will able to analyze to problems then transform in the model and type of mathematics equation. If the problems have available become mathematics equation, then calculate comprehend is needed. It is as not the absolute’s thing, because there are many count mountings tools today such as calculator and computer. So, the right myth is mathematics is always related to comprehension and reasoning.
The fourth myth, mathematics is abstract science and not related to reality. This myth is wrong, because the fact indicate if mathematics is realistic. It means, mathematics is analogy’s form from daily reality. The simple sample is the solution of Leonhard Euler, France mathematician, to Konisberg Bridge’s problem. Beside that, almost in all of the sectors, technology, economics, and social, mathematics play role significantly. Smart unimaginative person who able think is contained with program which is called expert system which have basic of Mathematics Fuzzy’s concept. Aerodynamic’s plane quantification and GPS concept also have basic of mathematics model concept, goniometry, and calculus. Almost all of economy’s theories and modern banking create by mathematics.
Whereas the fifth myth mention, mathematics is science which is boring, ungainly, and not interesting. This opinion is wrong. Although the mathematics solution felt exact because it has one solution, it not means that mathematics is ungainly and boring. Although it has one solution, we can solve the mathematics problems by various methods.
For example, to get solutions from two equations can use 3 method, are substitution method, elimination method, and curve. Another example, to proof the true of Pythagoras theorem, we can use any methods. Moreover according to mathematician, Bana G. Kartasasmita, there are 17 methods to proof Pythagoras method, today. The mathematics’s solution which has one characteristic cause pleasure to occur because it is explicit and exact.
Beside not boring, mathematics also ‘rekreatif’ and interesting. Albert Einstein, the greatest physics in 20 century, explain that mathematics is his prime weapon to formulate his relativists’s concept which had famous. According to Einstein, he like mathematics when his uncle explained to him the work procedure of mathematics like detective method, it is an act who he loves since he was children.
The mathematics methods are like a game. Beginning we have to identified variables or parameters by them attributes. After it, do the operation between those variable and parameter. The most please, when we do this operation, we can use trick freely until we get a correct solution. Freedom do tricks in this mathematics operations is challenging and invite enjoyable itself, it is like playing a game or adventure. Because of it, it is not be amazing if sometimes we find students who occupied be alone with them mathematics problems.
Beside it, the mathematics also have a number is called integer which is contains of mystery that very preoccupy. Such as you do the multiplying or additional amount to certain two numbers, then sometimes will appear a number which have certain symmetry form. The other example, you can indicate guess expertness a number correctly which has operable. For the people who comprehend mathematics yet, your ability to guess a number is reputed if it is a magic, whereas it is an operation.
Mathematics is science which is easy and enjoyable. Because of it, everyone be able to study mathematics diligently. So, the our prime assignment is bring down lose myths in the surrounding of mathematics.
- translet Indonesia-Inggris
Five lose myth on the subject of mathematics
A lot of lose myths about mathematics. This lose myths give great contribution in make partly of people allergy dislike mathematics. Finally, majority of our students got bad score in this subject, not because of disable, but because of early they feel allergy and be afraid, so they never or lazy to study this subject. Although there are many lose myths about mathematics, but there are five lose myths that have grew on and create bad perception to mathematics.
The first myth, mathematics is science that difficult until there are a little people who capable to comprehend by minimal IQ certain. It is mislead of course. Although it is not the easiest science, actually mathematics is a science that more easy than other science. For example, observe this problem comparison for the students of private elementary school grade six. The first problem is “Mention 3 Central Kalimantan traditional dances.” Second problem is “A circle is divided become 3 sections with comparison of center angle are 2:3:4, then count the value of center angle of the sections.”
Be in fact, the percentage of students who have right answer in second problem more than the percentage of students who have right answer in first problem. Without cause controvertion to occur, the samples above indicate if mathematics is not difficult science. The students regard as mathematics problems is difficult because they not comprehend numeral concept and measure concept correctly when they were elementary school students. If the numeral concept and the measure concept is governable, so analyzing and counting become easy and enjoyable.
The second myth, mathematics is memorized science from the many much formula. This myth made students lazy to study mathematics and finally not understand about mathematics. Whereas, substantively the mathematics is not the formula’s memorized science, because without comprehend the concept, the formulas that have memorized not useful. For example, there is a problem, “The machine is assembled by Benny longer 6 hour than Ahmad. If they can assemble the machine together in 4 hour, how long time needed by Ahmad to assemble the machine?”
Someone who memories quadrate equation will not able to answer those problem if they not capable take it to quadrate equation’s type. Actually, only a little mathematics’s formula needed to remind, whereas much of the other formulas not necessary remind, but it is comprehend enough. The example, if students understand the anatomy concept form cone slice, so more than 90% the cone slice’s formulas not remind.
The third myth, mathematics is always related to speed of calculate. The calculating belong mathematics, especially in the elementary school. But, capability of calculate quickly is not the most important in mathematics. The most important is concept comprehend. By concept comprehend, we will able to analyze to problems then transform in the model and type of mathematics equation. If the problems have available become mathematics equation, then calculate comprehend is needed. It is as not the absolute’s thing, because there are many count mountings tools today such as calculator and computer. So, the right myth is mathematics is always related to comprehension and reasoning.
The fourth myth, mathematics is abstract science and not related to reality. This myth is wrong, because the fact indicate if mathematics is realistic. It means, mathematics is analogy’s form from daily reality. The simple sample is the solution of Leonhard Euler, France mathematician, to Konisberg Bridge’s problem. Beside that, almost in all of the sectors, technology, economics, and social, mathematics play role significantly. Smart unimaginative person who able think is contained with program which is called expert system which have basic of Mathematics Fuzzy’s concept. Aerodynamic’s plane quantification and GPS concept also have basic of mathematics model concept, goniometry, and calculus. Almost all of economy’s theories and modern banking create by mathematics.
Whereas the fifth myth mention, mathematics is science which is boring, ungainly, and not interesting. This opinion is wrong. Although the mathematics solution felt exact because it has one solution, it not means that mathematics is ungainly and boring. Although it has one solution, we can solve the mathematics problems by various methods.
For example, to get solutions from two equations can use 3 method, are substitution method, elimination method, and curve. Another example, to proof the true of Pythagoras theorem, we can use any methods. Moreover according to mathematician, Bana G. Kartasasmita, there are 17 methods to proof Pythagoras method, today. The mathematics’s solution which has one characteristic cause pleasure to occur because it is explicit and exact.
Beside not boring, mathematics also ‘rekreatif’ and interesting. Albert Einstein, the greatest physics in 20 century, explain that mathematics is his prime weapon to formulate his relativists’s concept which had famous. According to Einstein, he like mathematics when his uncle explained to him the work procedure of mathematics like detective method, it is an act who he loves since he was children.
The mathematics methods are like a game. Beginning we have to identified variables or parameters by them attributes. After it, do the operation between those variable and parameter. The most please, when we do this operation, we can use trick freely until we get a correct solution. Freedom do tricks in this mathematics operations is challenging and invite enjoyable itself, it is like playing a game or adventure. Because of it, it is not be amazing if sometimes we find students who occupied be alone with them mathematics problems.
Beside it, the mathematics also have a number is called integer which is contains of mystery that very preoccupy. Such as you do the multiplying or additional amount to certain two numbers, then sometimes will appear a number which have certain symmetry form. The other example, you can indicate guess expertness a number correctly which has operable. For the people who comprehend mathematics yet, your ability to guess a number is reputed if it is a magic, whereas it is an operation.
Mathematics is science which is easy and enjoyable. Because of it, everyone be able to study mathematics diligently. So, the our prime assignment is bring down lose myths in the surrounding of mathematics.
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