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QveST [7]
3 years ago
10

Help me pleaseeeeee!!!

Mathematics
1 answer:
Svetlanka [38]3 years ago
3 0

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

\tan(x)  =  \frac{12}{29}  \\

\tan(x)  = 0.413

x≈ 22.475

x ≈22.5

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

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SOLVE THE QUESTION BELOW ASAP
qwelly [4]

Answer:

Part A) The graph in the attached figure (see the explanation)

Part B) 16 feet

Part C) see the explanation

Step-by-step explanation:

Part A) Graph the function

Let

h(t) ----> the height in feet of the ball above the ground

t -----> the time in seconds

we have    

h(t)=-16t^{2}+98

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex is a maximum

To graph the parabola, find the vertex, the intercepts,  and the axis of symmetry

<em>Find the vertex</em>

The function is written in vertex form

so

The vertex is the point (0,98)

Find the y-intercept

The y-intercept is the value of the function when the value of t is equal to zero

For t=0

h(t)=-16(0)^{2}+98

h(0)=98

The y-intercept is the point (0,98)

Find the t-intercepts

The t-intercepts are the values of t when the value of the function is equal to zero

For h(t)=0

-16t^{2}+98=0

t^{2}=\frac{98}{16}

square root both sides

t=\pm\frac{\sqrt{98}}{4}

t=\pm7\frac{\sqrt{2}}{4}

therefore

The t-intercepts are

(-7\frac{\sqrt{2}}{4},0), (7\frac{\sqrt{2}}{4},0)

(-2.475,0), (2.475,0)

Find the axis of symmetry

The equation of the axis of symmetry in a vertical parabola is equal to the x-coordinate of the vertex

so

x=0 ----> the y-axis

To graph the parabola, plot the given points and connect them

we have

The vertex is the point (0,98)

The y-intercept is the point (0,98)

The t-intercepts are (-2.475,0), (2.475,0)

The axis of symmetry is the y-axis

The graph in the attached figure

Part B) How far is the artifact fallen from the time t=0 to time t=1

we know that

For t=0

h(t)=-16(0)^{2}+98

h(0)=98\ ft

For t=1

h(t)=-16(1)^{2}+98

h(1)=82\ ft

Find the difference

98\ ft-82\ ft=16\ ft

Part C) Does the artifact fall the same distance from time t=1 to time t=2 as it does from the time t=0 to time t=1?

we know that

For t=1

h(t)=-16(1)^{2}+98

h(1)=82\ ft

For t=2

h(t)=-16(2)^{2}+98

h(2)=34\ ft

Find the difference

82\ ft-34\ ft=48\ ft

so

The artifact fall 48 feet from time t=1 to time t=2 and fall 16 feet from time t=0 to time t=1

therefore

The distance traveled from t=1 to t=2 is greater than the distance traveled from  t=0 to t=1

8 0
3 years ago
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