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Nutka1998 [239]
3 years ago
9

A ball is thrown into the air with an upward velocity of 48 ft/s. Its height h in feet after t seconds is given by the function

h = -16t^2 + 48t + 6. How long does it take the ball to reach its maximum height? What is the ball's maximum height?
Mathematics
1 answer:
Mila [183]3 years ago
6 0
So you have to find the velocity (or first derivative) from the position function of h=-16t^2+48t+6

the velocity is -32t+48 which you then have to set equal to 0 in order to find t
-32t+48=0
-32t=-48
t=1.5

you then take your t value and sub it back into your original position function
-16(1.5)^2+48(1.5)+6 which will give you a maximum height of 42 feet

so the time it takes is 1.5 seconds to reach a maximum height of 42 feet
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Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

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c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

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Answer:

Pretty sure it is C

Step-by-step explanation:

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