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Scrat [10]
3 years ago
6

I have no clue to what I even need to do.

Mathematics
1 answer:
Gre4nikov [31]3 years ago
4 0

\bf \textit{arc's length}\\\\ s=r\theta ~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad radians\\ \cline{1-1} \theta =\frac{3\pi }{2}\\ s=\frac{5\pi }{2} \end{cases}\implies \cfrac{5\pi }{2}=r\cdot \cfrac{3\pi }{2}

\bf \cfrac{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{3~~\begin{matrix} \pi \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{5~~\begin{matrix} \pi \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=r\implies \cfrac{5}{3}=r

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Paraphin [41]

Answer:

61.

Step-by-step explanation:

because 100-39=61 and 61+39=100!

3 0
4 years ago
Can sum one help and pls show work I don’t understand
makkiz [27]

Answer:

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Step-by-step explanation:

so you pick 2 sets of points I'm going to do (1,3) and (2,6) so that will be 6-3/2-1 which leaves you with 3/1x which is going to be 3x and that as a line will be y=3x+0 or y=3x because the y intercept is 0 and the slope is 3x so to graph that you all you have to do is plot the points in the table on to the graph or you go to (0,0) and go up 3 and to the right 1 and down 3 and left 1 but id rather plot the points. and that's all you have to do.

5 0
3 years ago
PLEASE HELP!!! WILL MARK BRAINLIEST!! THX! A golfer hits a ball with an initial velocity of 32.7 m/s from the ground. Find the f
OLEGan [10]

Answer:

See below

Step-by-step explanation:

<u>First Problem</u>

The ball hits the ground when h(t)=0, therefore:

h(t)=-4.9t^2+v_0t+h_0

0=-4.9t^2+32.7t

0=t(-4.9t+32.7)

t=0 and t=\frac{32.7}{4.9}\approx6.67

Since the ball is in the air before it hits the ground, t=6.67 (seconds) is the more appropriate choice.

<u>Second Problem</u>

The maximum height of the ball is determined when t=-\frac{b}{2a}, therefore:

t=-\frac{b}{2a}

t=-\frac{32.7}{2(-4.9)}

t=-\frac{32.7}{-9.8}

t\approx3.34

This means that the height of the ball is at its maximum after 3.34 seconds:

h(t)=-4.9t^2+32.7t

h(3.34)=-4.9(3.34)^2+32.7(3.34)

h(3.34)\approx54.55

Thus, the answer is 54.55 (meters).

<u>Third Problem</u>

Refer to the second problem

<u>Fourth Problem</u>

<u />h(t)=-4.9t^2+32.7t<u />

<u />h(4.3)=-4.9(4.3)^2+32.7(4.3)<u />

<u />h(4.3)\approx50.01<u />

<u />

Therefore, the height of the ball after 4.3 seconds is 50.01 (meters).

<u>Fifth Problem</u>

The ball will be 24 meters off the ground when h(t)=24, therefore:

h(t)=-4.9t^2+32.7t

24=-4.9t^2+32.7t

0=-4.9t^2+32.7t-24

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

t=\frac{-32.7\pm\sqrt{(32.7)^2-4(-4.9)(-24)}}{2(-4.9)}

t_1\approx0.84

t_2\approx5.83

Therefore, the ball will be 24 meters off the ground after 0.84 (seconds) and 5.83 (seconds)

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2 years ago
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evablogger [386]

Answer:

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Step-by-step explanation:

After adding like terms we would get 10^4 *10

Then we use the exponent rule and get 10^1^+^4

Which after adding would result in 10^5

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Answer:

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