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The exponent of a in 8a5 is 5 and the exponent of a in 2a3 is 3; 5 – 3 = 2. The exponent of a in our answer is 2. (8a5) ÷ (2a3) = 4a2. Now, let’s see an example where the dividend has a base that the divisor does not have, and the divisor has a base that the dividend does not have. Example (35g10) ÷ (5y4) = Divide the coefficient of the dividend by the coefficient of the divisor: 35 ÷ 5 = 7. Carry the bases of the dividend into the answer with its exponent, since that base is not present in the divisor.

4. The term 8m has a coefficient of 8, a base of m, and an exponent of 1. 5. The term 29 c7 has a coefficient of 29 , since c7 is multiplied by 29. The base is c and the exponent is 7. qxd:JSB 28 12/18/08 11:45 AM Page 28 algebra basics Practice 3 1. 3p and 3q have different bases, so they are unlike terms. 2. –k6 and 12k6 both have a base of k with an exponent of 6, so these terms are like terms. 3. 38 c2 and 8c2 both have a base of c with an exponent of 2, so these terms are like terms. 4. 8m4 and 8n4 have the same exponent, but they have different bases, so they are unlike terms.

Subtract: 16 – 20 = –4. 9. Replace b with –2: (2(–2) + 1)2 There is multiplication and addition inside the parentheses. Multiply first: 2(–2) = –4 The expression becomes (–4 + 1)2. Because the addition is in parentheses, it must be done before the exponent is handled: –4 + 1 = –3 We are left with (–3)2. Finally, square –3: (–3)2 = 9. 10. Replace a with 3: –5(3(3)2 – 24) There is multiplication, an exponent, and subtraction inside the parentheses. Because exponents come before multiplication and subtraction, handle the exponent first: (3)2 = 9 The expression becomes –5(3(9) – 24).

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