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Andrews [41]
3 years ago
8

In a circle an arc length 10 is intercepted by a central angle of 2/3 radius

Mathematics
1 answer:
djverab [1.8K]3 years ago
5 0

We have been given that in a circle an arc length 10 is intercepted by a central angle of 2/3. We are supposed to find the radius of the circle.

We will use arc-length formula to solve our given problem.

l=r\theta, where,

l = Arc length,

r = Radius,

\theta = Central angle corresponding to arc length.

Upon substituting our given values in arc-length formula, we will get:

10=r\cdot \frac{2}{3}

10\cdot \frac{3}{2}=r\cdot \frac{2}{3}\cdot \frac{3}{2}

5\cdot \frac{3}{1}=r

15=r

Therefore, the radius of the given circle would be 15 units.

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The triangle has side lengths 7, 10, and 12. Is it a right triangle? Explain your reasoning.
garri49 [273]

Answer:

Because triangle has three sides

3 0
3 years ago
PLZ HELP ASAP!!!
ozzi

Answer:

Part 1: Angle = 36°

Part 2: Angle = 72°

Step-by-step explanation:

If we have an angle "x", then

Its complement is 90 - x, and

supplement is 180 - x

Let's write equations and solve.

<u>Part 1:</u>

Complement = 1.5 times angle

90 - x = 1.5x

90 = 1.5x + x

90 = 2.5 x

x = 90/2.5 = 36

<u>Part 2:</u>

Now using 2nd equation:

3 * angle = 2 * Supplement

3x = 2 (180 - x)

3x = 360 - 2x

3x + 2x = 360

5x = 360

x = 360/5 = 72

7 0
3 years ago
Compare 117.93 and 117.859
Licemer1 [7]
117.93 is greater than 117.859 by .035.
3 0
3 years ago
544ml^2 is how many liters<br><br>show your work
ivolga24 [154]

Answer:

295.936

Step-by-step explanation:

First find out what 544^2 is.  That is 295,936.  There are 100 ml in 1 liter.  So divide 295,936 by 100 to see how many liters are in 295,936 ml.  

295,936÷100=295.936

7 0
3 years ago
The intensity of a light source at a distance is directly proportional to the strength of the source and inversely proportional
jolli1 [7]

Answer:

x=\frac{16}{\sqrt[3]{2}+1 }

Step-by-step explanation:

Q= illumination

I = intensity

Q= I/d^2

Q_total = \frac{I_1}{d_1^2}+\frac{I_2}{d_2^2}

= \frac{I}{x^2}+\frac{2I}{(16-x)^2}

now Q' = 0

⇒I{-\frac{2}{x^3}}+\frac{4}{(16-x)^3}

x=\frac{16}{\sqrt[3]{2}+1 }

[/tex]\frac{1}{x^3} = \frac{2}{(16-x)^3}

x=\frac{16}{\sqrt[3]{2}+1 }

is the required point

5 0
3 years ago
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