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McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
5. Using the Distributive Property
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Exercise 48 Page 41

Practice makes perfect
a

Let's factor the right hand side of the given formula.

- 16t^2-96t+160
SplitIntoFactors

Split into factors

16 * (- t^2) - ( 16)* 6t+ ( 16)* (10)
FactorOut

Factor out the GCF

16(- t^2-6t+10)

This expression represents the height in factored form.

b

To find the initial height, we need to substitute 0 for t into the given equation.

h=- 16t^2-96t+160
Substitute

t= 0

h=- 16( 0)^2-96( 0)+160
â–¼
Evaluate right-hand side
CalcPow

Calculate power

h=- 16(0)-96(0)+160
IdPropMult

Identity Property of Multiplication

h=0-0+160
IdPropAdd

Identity Property of Addition

h=160

The rider is dropped from 160 ft.

c

Let's substitute 3 for t.

h=- 16t^2-96t+160
Substitute

t= 3

h=- 16( 3)^2-96( 3)+160
â–¼
Evaluate right-hand side
CalcPow

Calculate power

h=- 16(9)-96(3)+160
Multiply

Multiply

h=- 144-288+160
AddSubTerms

Add and subtract terms

h=- 272

The model shows that the rider will be at - 272 ft height. However, this is not possible since the height cannot be a negative value as this would imply that the rider would be below the ground.

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