BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 1.2, Problem 42E

(a)

To determine

To simplify: given expression b4(3ab3)(2a2b5) .

Expert Solution

Answer to Problem 42E

  b4(3ab3)(2a2b5)=6a3b2

Explanation of Solution

Given information:

Given expression

  b4(3ab3)(2a2b5)

Calculation:

Consider the given expression.

  b4(3ab3)(2a2b5)

And eliminate any negative exponents. To simplify this, first rewrite this expression as a single product.

  b4(3ab3)(2a2b5)=32aa2b4b3b5=6aa2b4b3b5

Now, use the first law of exponents, which states

  aman=am+n

And combine like terms

  6aa2b4b3b5=6a1+2b4+35=6a3b2

Hence,

  b4(3ab3)(2a2b5)=6a3b2

(b)

To determine

To evaluate: the given expression (2s3t2)(14s7t)(16t4) .

Expert Solution

Answer to Problem 42E

  (2s3t2)(14s7t)(16t4)=8s10t3

Explanation of Solution

Given information:

Given expression

  (2s3t2)(14s7t)(16t4)

Calculation:

Consider the given expression.

  (2s3t2)(14s7t)(16t4)

And eliminate any negative exponents. To simplify this, first rewrite this expression as a single product.

  (2s3t2)(14s7t)(16t4)=21416s3s7t2tt4=8s3s7t2tt4

Now, use the first law of exponents, which states

  aman=am+n

And combine like terms

  8s3s7t2tt4=8s3+7t2+1+4=8s10t3

Hence,

  (2s3t2)(14s7t)(16t4)=8s10t3

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