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kvv77 [185]
3 years ago
12

Colton bounces a ball 3.268 feet infront of his feet. The path of the ball from the time it hits the ground until it lands on th

e floor is represented by
Mathematics
1 answer:
Otrada [13]3 years ago
5 0

here is the complete and correct question. Colton bounces a ball 3.268 feet in front of his feet. The path of

the ball from the time it hits the ground until it lands on the floor is represented by

f(x) =-4(x - 5)2 + 12

Assuming that Colton's feet are located at the origin, (0, 0), what is the maximum height of the ball (in feet)?

Answer:

12ft

step by step explanation:

This equation has been written in vertex form, and it has its vertex at (5, 12). Because the scale factor is not positive, it has a negative value, the graph for the question is going to open downward, such that the vertex,

that is the maximum height is 12 ft.

we have The equation of a parabola with vertex (h, k) to be

f(x) = a(x -h)² + k

when put in Comparison to your question function, we get that

a=-4

h=5

k=12

so the vertex (h, k) = (5, 12).

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3 years ago
Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me​
o-na [289]

Answer:

\frac{3}{2}

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

                   = - cos60cosx - sin60sinx

                   = - \frac{1}{2} cosx - \frac{\sqrt{3} }{2} sinx

squaring to obtain cos² (120 + x)

= \frac{1}{4}cos²x + \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

                   = -cos60cosx + sin60sinx

                   = - \frac{1}{2}cosx + \frac{\sqrt{3} }{2}sinx

squaring to obtain cos²(120 - x)

= \frac{1}{4}cos²x - \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + \frac{1}{4}cos²x + \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x + \frac{1}{4}cos²x - \frac{\sqrt{3} }{2}sinxcosx + \frac{3}{4}sin²x

= cos²x + \frac{1}{2}cos²x + \frac{3}{2}sin²x

= \frac{3}{2}cos²x + \frac{3}{2}sin²x

= \frac{3}{2}(cos²x + sin²x) = \frac{3}{2}

                 

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