By Bob Miller

With Bob Miller at your facet, you by no means must be clueless approximately math again!Algebra and calculus are difficult on highschool scholars such as you. Professor Bob Miller, with greater than 30 years' instructing adventure, is a grasp at making the complicated easy, and his now-classic sequence of Clueless learn aids has helped tens of millions comprehend the harsh subjects.Bob Miller expands at the prior variation for simple arithmetic and pre-algebra scholars, with new fabric designed to spice up realizing and grades. Miller breaks open the more durable parts of arithmetic so that you can grasp the fabrics and ace each attempt.

**Read Online or Download Bob Miller's Basic Math and Pre-Algebra for the Clueless PDF**

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**Additional resources for Bob Miller's Basic Math and Pre-Algebra for the Clueless**

**Example text**

N OTE EXAMPLE 17— (−3a2b4)3 = (−3)3(a2)3(b4)3 Answer: −27a6b12. Rememn ber: n n (xy) = x y . EXAMPLE 18— −2x 4 5 ᎏ . 3y6 Top: (−2)5 = −32; (x4)5 = x20. Bottom: 3 = 243; (y ) = y . 5 −32x 20 Answer ᎏ 243y 30 6 5 30 Note: 5th and last law of exponents: a n an ᎏᎏ = ᎏn b b 24 B O B M I L L E R ’ S B A S I C M AT H A N D P R E A L G E B R A EXAMPLE 19— . Simplify inside ( ) first. ᎏ 3a b 12a4b5 2 9 2 12 a4 b5 = ᎏ ᎏ9 ᎏ2 3 a b 2 =ᎏ a 4b3 = ᎏ a5 2 16b6 10 EXAMPLE 20— (4a3b6)2 Outside exponents are different; they must ᎏ .

Exponents are subtracted; base stays 1120 1115 the same. Short Division This is the opposite of adding fractions, which says a b a+b 2 3 5 ᎏ + ᎏ = ᎏ orrr ᎏ + ᎏ = ᎏ . c c c 7 7 7 a+b a b The opposite is ᎏ = ᎏ + ᎏ . c c c EXAMPLE 23— 55a9 + 35a7 55a9 35a7 ᎏᎏ =ᎏ +ᎏ = 11a4 + 7a2. 5a5 5a5 5a5 EXAMPLE 24— 16a8 − 18a7 + 4a3 + 2a ᎏᎏᎏ . 4a3 16a8 18a7 4a3 9a4 1 2a ᎏ −ᎏ + ᎏ3 + ᎏ3 = 4a5 − ᎏ + 1 + ᎏ2 3 3 4a 4a 4a 2 4a 2a 1. The “1” must be the third term because all the terms are added (anything except 0 over itself is 1).

From 2 to 1 is down 1: What we just showed is a negative times a negative is a positive. ) More generally, we need to look at only negative signs in multiplication and division problems. Odd number of negative signs, answer is negative. Even number of negative signs, answer is positive. E X A M P L E 1 3 ( V E RY I M P O R TA N T ) — A. −32; B. (−3)2; C. −(−3)2. The answer in each case is 9. ” The answer is the number of minus signs. A. −32 = −(3 × 3) = −9 B. (−3)2 = (−3)(−3) = +9 C. −(−3)2 = −(−3)(−3) = −9 One minus sign Two minus signs Three minus signs We need to explain a little.