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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

Chapter Review

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Exercise 22 Page 286

To evaluate h(x), substitute the given values into the equation.

h(2)=53
h(7)=33

Practice makes perfect

To find the value of h(x)=-4x+61 when x=2 and x=7, we will substitute the values for x and evaluate. Let's start with x= 2.

h(x)=-4x+61

x= 2

h( 2)=-4( 2)+61

(- a)b = - ab

h(2)=-8+61

Add terms

h(2)=53

Therefore, when x=2, h(x)=53. Now let's try x=7.

h(x)=-4x+61

x= 7

h( 7)=-4( 7)+61

(- a)b = - ab

h(7)=-28+61

Add terms

h(7)=33

When x=7, h(x)=33.