Mathematics, 18.03.2020 17:22 aurora32

The equation h(t) = -16t² + 80t + 64 represented the height, in feet, of a potato t seconds after it has been launched.
a. Write an equation that can be solved to find when the potato hits the ground. Then solve the equation.
b. Write an equation that can be solved to find when the potato is 40 feet off the ground. Then solve the equation

Answers: 2

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Answer from: gomez36495983

The potato will hit the ground after 5.6 secsThe potato will be 40 feet above the ground after 10.3 secs

Given the equation that models the height of the potato given as h(t) = -16t² + 80t + 64, the potato hits the ground at when h(t) = 0

The equation becomes -16t² + 80t + 64 = 0

Factorizing there result:

16t² - 80t - 64 = 0

t² - 5t - 4 = 0

On factorizing the value of t

t = 5±√25-4(-4)/2

t =5±√25+16/2

t=5±√41/2

t = 5+6.4/2 and 5 - 6.4/2

t = 5.7secs

Hence the potato will hit the ground after 5.6 secs

b) If the potato is 40 feet off the ground, the equation becomes

-16t² + 80t + 64 = 40

Factorizing there result:

16t² - 80t - 24 = 0

2t² - 10t - 3 = 0

On factorizing the value of t

t = 10±√100-4(-3)/2

t =10±√100+12/2

t=10±√112/2

t = 10+10.6/2 and 10 - 10.6/2

t = 10.3 secs

The potato will be 40 feet above the ground after 10.3 secs

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Answer from: Jasten

Part A) The potato hit the ground at t=5.70 seconds (see the explanation)

Part B) The potato is 40 feet off the ground at the time t=5.28 seconds (see the explanation)

Step-by-step explanation:

we have

h(t)=-16t^2+80t+64

where

h(t) is the height of a potato in feet

t is the time in seconds

Part A) Write an equation that can be solved to find when the potato hits the ground. Then solve the equation

we know that

When the potato hit the ground, the value of h(t) must be equal to zero

For h(t)=0

-16t^2+80t+64=0

Solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

-16t^2+80t+64=0

$a=-16\\b=80\\c=64$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(64)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{10,496}} {-32}$

$t=\frac{-80+\sqrt{10,496}} {-32}=-0.70$

$t=\frac{-80-\sqrt{10,496}} {-32}=5.70$

therefore

The potato hit the ground at t=5.70 seconds

Part B) Write an equation that can be solved to find when the potato is 40 feet off the ground. Then solve the equation

For h(t)=40 ft

substitute in the quadratic equation

-16t^2+80t+64=40

-16t^2+80t+24=0

Solve the quadratic equation

we have

$a=-16\\b=80\\c=24$

substitute in the formula

$t=\frac{-80\pm\sqrt{80^{2}-4(-16)(24)}} {2(-16)}$

$t=\frac{-80\pm\sqrt{7,936}} {-32}$

$t=\frac{-80+\sqrt{7,936}} {-32}=-0.28$

$t=\frac{-80-\sqrt{7,936}} {-32}=5.28$

therefore

The potato is 40 feet off the ground at the time t=5.28 seconds

Answer from: Quest

1. 2

2. 8

3. 9

step-by-step explanation:

lol can i have brainliest, i spent hours trying to figure these mind boggling math questions out

Answer from: Quest

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4

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