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defon
3 years ago
9

A ball is thrown into the air. The height h, in feet, of the ball after x seconds is given by the function h = −16(x − 2)2 + 72.

What is the equation in standard form and the maximum height of the ball?
A. h(x) = −16x2 + 32x + 72; 72 ft
B. h(x) = −16x2 − 32x + 72; 32 ft
C. h(x) = −16x2 − 64x + 32; 32 ft
D. h(x) = −16x2 + 64x + 8; 72 ft
Mathematics
2 answers:
Vinil7 [7]3 years ago
5 0

Answer:

A. -16x²+64x+8

B. 72 ft.

Step-by-step explanation: A. Expand the perfect square binomial -16(x²-4x+4)+72, dist. -16 (-16x²+64x-64+72), then combine like terms (-16x²+64x+8)

B. Find the abs. max. by looking back at the vertex form. The max is (2,72). Use the y-value for the height.

Have a great day : )

Annette [7]3 years ago
4 0

Answer:

Step-by-step explanation:

h = −16(x − 2)² + 72

 =  −16((x − 2)(x − 2)) + 72                   expand the square  (x - 2)² =  (x - 2)(x - x)

 =  -16 ((x)(x) + (x)(-2) + (-2)(x) + (-2)(-2)) + 72

  = -16 (x² + (-2x) + (-2x) + (4) ) + 72

  = -16 (x² + (-4x)  + (4) ) + 72

  = -16 (x² - 4x  + 4 ) + 72

  = (-16)(x²) + (-16)(-4x) +(-16)(4) + 72

  =   -16x²  + 64x  - 64 + 72

  =   -16x²  + 64x  + 8              My guess is D   ; 72 ???

I have no clue what the  ; 72 means after the given answer.  It must be a standard form thing I never learned.

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Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

f(x) = 2 is a valid solution.

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Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need A= 2, and at that point the first condition is guaranteed; using the second to find k we get 2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi

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Sky is the limit.

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6 0
3 years ago
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