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Mathematics

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Ber [7]3 years ago

4 0

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Step-by-step explanation:

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A juggler tosses a ball into the air . The balls height, h and time t seconds can be represented by the equation h(t)= -16t^2+40

malfutka [58]

PART A

The given equation is

$h(t) = - 16 {t}^{2} + 40t + 4$

In order to find the maximum height, we write the function in the vertex form.

We factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + 4$

We add and subtract

$- 16(- \frac{5}{4} )^{2}$

to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t) + - 16( - \frac{5}{4})^{2} - -16( - \frac{5}{4})^{2} + 4$

We again factor -16 out of the first two terms to get,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4})^{2} ) - -16( - \frac{5}{4})^{2} + 4$

This implies that,

$h(t) = - 16 ({t}^{2} - \frac{5}{2} t + ( - \frac{5}{4}) ^{2} ) + 16( \frac{25}{16}) + 4$

The quadratic trinomial above is a perfect square.

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +25+ 4$

This finally simplifies to,

$h(t) = - 16 ( t- \frac{5}{4}) ^{2} +29$

The vertex of this function is

$V( \frac{5}{4} ,29)$

The y-value of the vertex is the maximum value.

Therefore the maximum value is,

$29$

PART B

When the ball hits the ground,

$h(t) = 0$

This implies that,

$- 16 ( t- \frac{5}{4}) ^{2} +29 = 0$

We add -29 to both sides to get,

$- 16 ( t- \frac{5}{4}) ^{2} = - 29$

This implies that,

$( t- \frac{5}{4}) ^{2} = \frac{29}{16}$

$t- \frac{5}{4} = \pm \sqrt{ \frac{29}{16} }$

$t = \frac{5}{4} \pm \frac{ \sqrt{29} }{4}$

$t = \frac{ 5 + \sqrt{29} }{4} = 2.60$

or

$t = \frac{ 5 - \sqrt{29} }{4} = - 0.10$

Since time cannot be negative, we discard the negative value and pick,

$t = 2.60s$

8 0

3 years ago

Click or touch the graph to plot the points.

lina2011 [118]

Answer:

the answer is C

Step-by-step explanation:

the y intercept is on the right, so (x, y)

and the third option has (4,8) so 8 is on the y

4 0

3 years ago

Let g(x)=2x and h(x)=x^2+4 find the value. (hog)(a)

lisov135 [29]

In short, (h o g)(a) is just h( g(a) ).

so what we can do is simply get g(a) first and then plug that in h(x).

$\bf \begin{cases}
g(x)&=2x\\
h(x)&=x^2+4\\
(h\circ g)(a)&=h(~~g(a)~~)
\end{cases}
\\\\\\
g(a)=2(a)\implies g(a)=2a
\\\\\\
h(~~g(a)~~)\implies h(~~2a~~)=(2a)^2+4
\\\\\\
h(~~2a~~)=(2^2a^2)+4\implies h(~~2a~~)=4a^2+4$

8 0

3 years ago

What is equivalent to -3(x-9)

ivolga24 [154]

$-3(x-9)=-3\cdot x-3\cdot(-9)=-3x+27$