Exercise 13 Page 39

Practice makes perfect

We want to find when the spider's height is 0. To find it, we will substitute 0 for h into the given equation.

h=33.3t-16t^2

Substitute

h= 0

0=33.3t-16t^2

FactorOut

Factor out t

0=t(33.3-16t)

To solve this equation, we will apply the Zero Product Property.

0=t(33.3-16t)

ZeroProdProp

Use the Zero Product Property

lct=0 & (I) 33.3-16t=0 & (II)

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(II): Solve for t

SubEqn

(II): LHS-33.3=RHS-33.3

lct=0 & (I) - 16t=- 33.3 & (II)

DivEqn

(II): .LHS /(- 16).=.RHS /(- 16).

lct=0 & (I) t=- 33.3/- 16 & (II)

CalcQuot

(II): Calculate quotient

lct=0 & (I) t=2.08125 & (II)

The spider's height is 0 feet at 0 seconds and 2.08125 seconds.

To find the spider's height after 1 second, we need to substitute 1 for t in the equation.

h=33.3t-16t^2

Substitute

t= 1

h=33.3( 1)-16( 1)^2

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Simplify right-hand side

BaseOne

1^a=1

h=33.3(1)-16(1)

IdPropMult

Identity Property of Multiplication

h=33.3-16

SubTerm

Subtract term

h=17.3

The height of the spider is 17.3ft after 1 second. Let's find its height after 2 seconds.

h=33.3t-16t^2

Substitute

t= 2

h=33.3( 2)-16( 2)^2

▼

Simplify right-hand side

CalcPow

Calculate power

h=33.3(2)-16(4)

Multiply

h=66.6-64

SubTerm

Subtract term

h=2.6

The height of the spider is 2.6ft after 2 seconds.

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Exercises p.39-42

Exercise 13 Page 39

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