A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by ℎ = −16푡 ଶ+103푡+5 where t is time (in seconds).

Answers

Answer 1

Question:

A fastball is hit straight up over home plate. The ball's height, h (in feet), from the ground is modeled by h(t)=-16t^2+80t+5, where t is measured in seconds.  Write an equation to determine how long it will take for the ball to reach the ground.

Answer:

[tex]t = 5.0625[/tex]

Step-by-step explanation:

Given

[tex]h(t)=-16t^2+80t+5[/tex]

Required

Find t when the ball hits the ground

This implies that h(t) = 0

So, we have:

[tex]0=-16t^2+80t+5[/tex]

Reorder

[tex]-16t^2+80t+5 = 0[/tex]

Using quadratic formula, we have:

[tex]t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]

Where

[tex]a = -16[/tex]      [tex]b =80[/tex]      [tex]c = 5[/tex]

So, we have:

[tex]t = \frac{-80\±\sqrt{80^2 - 4*-16*5}}{2*-16}[/tex]

[tex]t = \frac{-80\±\sqrt{6400 +320}}{-32}[/tex]

[tex]t = \frac{-80\±\sqrt{6720}}{-32}[/tex]

[tex]t = \frac{-80\±82.0}{-32}[/tex]

This gives:

[tex]t = \frac{-80+82.0}{-32}[/tex] or [tex]t = \frac{-80-82.0}{-32}[/tex]

[tex]t = \frac{2}{-32}[/tex] or [tex]t = \frac{-162}{-32}[/tex]

[tex]t = -\frac{2}{32}[/tex] or [tex]t = \frac{162}{32}[/tex]

But time can not be negative.

So, we have:

[tex]t = \frac{162}{32}[/tex]

[tex]t = 5.0625[/tex]

Hence, time to hit the ground is 5.0625 seconds


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Answer:

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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Step 2: M for Multiply. You multiply your answer from step 1 and your divisor: .

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Step 1: D for Divide. ...

Step 2: M for Multiply. ...

Step 3: S for Subtract.

A(x) = 2 (3x - 2)(x - 5)

Answers

Answer:

Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

), where x  

1

​  

 and x  

2

​  

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

​  

 

​  

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

​  

 

​  

 

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

​  

 

​  

 

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

​  

 

​  

 

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

​  

 

​  

 

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

​  

 

​  

 

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

​  

 

​  

 

Multiply 2 times −1.

x=  

−2

3±  

29

​  

 

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is plus. Add 3 to  

29

​  

.

x=  

−2

29

​  

+3

​  

 

Divide 3+  

29

​  

 by −2.

x=  

2

−  

29

​  

−3

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is minus. Subtract  

29

​  

 from 3.

x=  

−2

3−  

29

​  

 

​  

 

Divide 3−  

29

​  

 by −2.

x=  

2

29

​  

−3

​  

 

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

). Substitute  

2

−3−  

29

​  

 

​  

 for x  

1

​  

 and  

2

−3+  

29

​  

 

​  

 for x  

2

​  

.

−x  

2

−3x+5=−(x−  

2

−  

29

​  

−3

​  

)(x−  

2

29

​  

−3

​  

)

EVALUATE

5−3x−x  

2Quadratic polynomial can be factored using the transformation ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

), where x  

1

​  

 and x  

2

​  

 are the solutions of the quadratic equation ax  

2

+bx+c=0.

−x  

2

−3x+5=0

All equations of the form ax  

2

+bx+c=0 can be solved using the quadratic formula:  

2a

−b±  

b  

2

−4ac

​  

 

​  

. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=  

2(−1)

−(−3)±  

(−3)  

2

−4(−1)×5

​  

 

​  

 

Square −3.

x=  

2(−1)

−(−3)±  

9−4(−1)×5

​  

 

​  

 

Multiply −4 times −1.

x=  

2(−1)

−(−3)±  

9+4×5

​  

 

​  

 

Multiply 4 times 5.

x=  

2(−1)

−(−3)±  

9+20

​  

 

​  

 

Add 9 to 20.

x=  

2(−1)

−(−3)±  

29

​  

 

​  

 

The opposite of −3 is 3.

x=  

2(−1)

3±  

29

​  

 

​  

 

Multiply 2 times −1.

x=  

−2

3±  

29

​  

 

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is plus. Add 3 to  

29

​  

.

x=  

−2

29

​  

+3

​  

 

Divide 3+  

29

​  

 by −2.

x=  

2

−  

29

​  

−3

​  

 

Now solve the equation x=  

−2

3±  

29

​  

 

​  

 when ± is minus. Subtract  

29

​  

 from 3.

x=  

−2

3−  

29

​  

 

​  

 

Divide 3−  

29

​  

 by −2.

x=  

2

29

​  

−3

​  

 

Factor the original expression using ax  

2

+bx+c=a(x−x  

1

​  

)(x−x  

2

​  

). Substitute  

2

−3−  

29

​  

 

​  

 for x  

1

​  

 and  

2

−3+  

29

​  

 

​  

 for x  

2

​  

.

−x  

2

−3x+5=−(x−  

2

−  

29

​  

−3

​  

)(x−  

2

29

​  

−3

​  

)

EVALUATE

5−3x−x  

2

Step-by-step explanation:

is this proportional relationship? please help!!

Answers

Answer:

yes, the slope is the same throughout the table

Step-by-step explanation:

Answer:

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Step-by-step explanation:

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