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**Evaluate the expression for the given value(s) of the variable(s).**

Example 4 Evaluating Algebraic Expressions Evaluate the expression for the given value(s) of the variable(s). 3x + 4, x = 7 a. – s = 2, b. p 4 10s 3 3p 2, SOLUTION 3x + 4 = 3 a. ( ) 7 Substitute 7 for x. = 21 + 4 Multiply. = 25 Add.

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**Example 4 b. 10s 3 – 3p 2 = 10 ( ) 3 2 – 3 ( ) 2 4 = 10 ( ) 8 – 3 16 =**

Evaluating Algebraic Expressions b. 10s 3 – 3p 2 = 10 ( ) 3 2 – 3 ( ) 2 4 Substitute 2 for s and 4 for p. Evaluate powers. = 10 ( ) 8 – 3 16 = 80 Multiply. – 48 = 32 Subtract.

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Example 5 Evaluating a Real-Life Expression WATERFALL A river carries a stone over a waterfall. The stone hits the water 5 seconds later. To find the height of the waterfall, use the expression 16t2, which gives the distance in feet that a dropped object falls in t seconds. SOLUTION 16t 2 = 16 ( ) 2 5 Substitute 5 for t. Evaluate the power. 16 ( ) 25 = = 400 Multiply.

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**The waterfall is about 400 feet high.**

Example 5 Evaluating a Real-Life Expression ANSWER The waterfall is about 400 feet high.

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**Evaluate the expression for the given value(s) of the variable(s).**

Guided Practice for Examples 4 and 5 Evaluate the expression for the given value(s) of the variable(s). 17. 4r + 6, r = 5 ANSWER 26 18. ( ), 5 – y 3 = 8 ANSWER 9 19. n2 + q3, n = 1, q 2 ANSWER 9

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**Guided Practice 20. WHAT IF?**

for Examples 4 and 5 20. WHAT IF? In Example 5, suppose the stone hits the water after 2 seconds. Find the height of the waterfall. ANSWER 64 ft

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EXAMPLE 1 Evaluate powers a. (–5) 4 b. –5 4 = (–5) (–5) (–5) (–5)= 625 = –(5 5 5 5)= –625.

EXAMPLE 1 Evaluate powers a. (–5) 4 b. –5 4 = (–5) (–5) (–5) (–5)= 625 = –(5 5 5 5)= –625.

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