BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter T, Problem 2ADT

(a)

To determine

To simplify: The expression 20032.

Expert Solution

Answer to Problem 2ADT

The expression 20032 is simplified as 62.

Explanation of Solution

Consider the expression 20032.

Observe that,

200=2×2×2×5×5=(2)2×(5)2×2=(2×5)2=102

Again observe that,

32=2×2×2×2×2=(2)2×(2)2×2=(2×2)2=42

Substitute 200=102 and 32=42 in 20032 and simplify the above expression as follows,

20032=10242=(104)2=62

Thus, the expression 20032 is simplified as 62.

(b)

To determine

To simplify: The expression (3a3b3)(4ab2)2.

Expert Solution

Answer to Problem 2ADT

The expression (3a3b3)(4ab2)2 is simplified as 48a5b7.

Explanation of Solution

Consider the expression (3a3b3)(4ab2)2.

Simplify the above expression as follows,

(3a3b3)(4ab2)2=(3a3b3)(16a2b4)=3×16a3+2b3+4=48a5b7

Thus, the expression (3a3b3)(4ab2)2 is simplified as 48a5b7.

(c)

To determine

To simplify: The expression (3x32y3x2y12)2.

Expert Solution

Answer to Problem 2ADT

The expression (3x32y3x2y12)2 is simplified as x9y7.

Explanation of Solution

Consider the expression (3x32y3x2y12)2.

Simplify the above expression as follows,

(3x32y3x2y12)2=(x2y123x32y3)2(a1=1a)=(x2y12)2(3x32y3)2=(x4y19x3y6)=(x49x3y7)

=x9y7

Thus, the expression (3x32y3x2y12)2 is simplified as x9y7.

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