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Pepsi [2]
2 years ago
13

A circle has a central angle measuring StartFraction 7 pi Over 10 EndFraction radians that intersects an arc of length 33 cm. Wh

at is the length of the radius of the circle? Round your answer to the nearest whole cm. Use 3. 14 for Pi. 11 cm 15 cm 22 cm 41 cm.
Mathematics
1 answer:
cupoosta [38]2 years ago
8 0

The radius of the given circle with a central angle of 7π/10 and arc length of 33cm is 15cm.

It is given that

The central angle of the given circle = 7π/10 radians

Length of arc = 33cm

<h3>What is the formula for arc length?</h3>

Arc length = central angle* radius

From the above formula

Radius = Arc length/Central Angle

Radius = \frac{33}{(7\pi)/10 }

Radius = 15cm

Therefore, radius of the given circle is 15cm.

to get more about circles visit:

brainly.com/question/22965557

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Solve for u. 20 − (u · 0.9) = 4.7
alina1380 [7]

Answer:

20 - 0.9u = 4.7

-0.9u = 4.7 - 20

-0.9u = -15.3

Divide both sides by - 0.9

u = 17

Hope this helps.

3 0
3 years ago
What would be a reasonable estimate of the value of 'm' in this equation
bixtya [17]
B. 188-80=108
The closest answer there is 100
5 0
3 years ago
X/7 = -3<br> 1. x = -21<br> 2. x = -10<br> 3. x = 4<br> 4. x = 21
SOVA2 [1]
X/7= -3
X= -21

Hope this helps!
5 0
3 years ago
Find the length of UC? Please help
Roman55 [17]

Answer:

The choose C. 18

Step-by-step explanation:

UC —> 105+82=187 —> 96+22+51=169 —> 187–169=18

I hope I helped you^_^

7 0
3 years ago
Which of the following is a solution of z^5 = 1 + √3 i?
nordsb [41]

Answer:

Option 2 is right

Step-by-step explanation:

Given that

z^5=1+\sqrt{3} i

We can write this in polar form with modulus and radius

|z^5|= \sqrt{1+3} =2\\tan of Angle t =\sqrt{3} \\

Hence angle = 60 degrees and

|z^5|= 2(cos60+isin60)

Since we have got 5 roots for z, we can write 60, 420, 780, etc. with periods of 360

Using Demoivre theorem we get 5th root would be

5th root of 2 multiplied by 1/5 th of 60, 420, 780,....

z= \sqrt[5]{2} (cos12+isin12)\\z=\sqrt[5]{2} (cos84+isin84)\\\\z=\sqrt[5]{2} (cos156+isin156)\\\\z=\sqrt[5]{2} (cos228+isin228)\\\\z=\sqrt[5]{2} (cos300+isin300)\\

Out of these only 2nd option suits our answer

Hence answer is Option 2.

8 0
3 years ago
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