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Lubov Fominskaja [6]
3 years ago
11

What is the length of the arc on a circle with radius 20 in. intercepted by a 15° angle? Use 3.14 for π . Round the answer to th

e hundredths place.
Mathematics
1 answer:
Fofino [41]3 years ago
5 0
We know, length of arc = Ф/360 * 2πr
= 15/360 * (2 * 3.14 * 20)
= 1/24 * (125.6)
= 5.23

In short, Your Answer would be 5.23 Inches

Hope this helps!
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The answer is 8 and 3 fourths which would be written as 8 3/4
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Zx and Zy are supplementary angles. Zy measures 108°,
gulaghasi [49]

Answer:

72°

Step-by-step explanation:

Given

Zy = 108°

Zx = ?

We know

Zx + Zy = 180° {being supplementary angles }

Zx + 108° = 180°

Zx = 180° - 108°

Zx = 72°

Hope it will help :)

4 0
3 years ago
Meg skateboards One-fourth of a mile in One-half of an hour. At this rate, how far will she skateboard in an hour?
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Answer: i'm pretty sure Meg would have skated 1/2 of a mile in an hour

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3 years ago
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Determine the domain of the function h=9x/x(x^2-49)
Juliette [100K]

ANSWER

( -  \infty , - 7) \cup( - 7 , 0) \cup(0 , 7 )\cup(7 , +   \infty )

EXPLANATION

The given function is

h(x) =  \frac{9x}{x( {x}^{2} - 49) }

This function is defined for values where the denominator is not equal to zero.

x( {x}^{2}  - 49) \ne0

x(x - 7)(x + 7) = 0

The domain is

x \ne - 7, x \ne0, \: and \:  x \ne 7,

Or

( -  \infty , - 7) \cup( - 7 , 0) \cup(0 , 7 )\cup(7 , +   \infty )

8 0
3 years ago
A parabola can be drawn given a focus of (-11, -2) and a directrix of x= -3
fgiga [73]

Check the picture below, so the parabola looks more or less like that.

now, the vertex is half-way between the focus point and the directrix, so that puts it where you see it in the picture, and the horizontal parabola is opening to the left-hand-side, meaning that the distance "P" is negative.

\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\supset}\qquad \stackrel{"p"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

\begin{cases} h=-7\\ k=-2\\ p=-4 \end{cases}\implies 4(-4)[x-(-7)]~~ = ~~[y-(-2)]^2 \\\\\\ -16(x+7)=(y+2)^2\implies x+7=-\cfrac{(y+2)^2}{16}\implies x=-\cfrac{1}{16}(y+2)^2-7

8 0
2 years ago
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