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3 years ago

In a circle, an angle measuring 2π radians intercepts an arc of length 14π. Find the radius of the circle in simplest form.

Mathematics

1 answer:

Stels [109]3 years ago

6 0

Answer:

7 radians

Step-by-step explanation:

In a circle, an angle measuring 2π radians intercepts an arc of length 14π. Find the radius of the circle in simplest form.

Since our Angle in in radians

The formula for Arc length = S = rθ

S = 14π

θ = central angle = 2π radians

Hence

r = S/θ

r = 14π/2π radians

r = 7 radians

Radius = 7 radians

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Two resistors, with resistances R1 and R2, are connected in series. R1 is normally distributed with mean 100 ohms and standard d

puteri [66]

Answer:

The probability that R2 is bigger than R1 is P=0.9633.

Step-by-step explanation:

We need to compute the probability that R2 is bigger than R1:

$P(R_2>R_1)$

To do this, we define a new variable Das the difference between the two resistances:

$D=R_2-R_1$

Then,

$P(R_2>R_1)=P(R_2-R_1>0)=P(D>0)$

As R1 and R2 are normal random variables, the properties of D are:

$\mu_D=\mu_{R2}-\mu_{R1}=120-100=20\\\\\sigma_D=\sqrt{\sigma_{R2}^2+\sigma_{R1}^2}=\sqrt{10^2+5^2}=\sqrt{100+25}=\sqrt{125}=11.18$

Then, we can calculate the z-value for D=0

$z=\frac{x-\mu}{\sigma} =\frac{0-20}{11.18}=\frac{-20}{11.18}= -1.79$

$P(D>0)=P(z>-1.79)=0.9633$

8 0

3 years ago

What the hell does any of this mean

Ksenya-84 [330]

Answer:

1/16x^4y^10

Step-by-step explanation:

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laiz [17]

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What is the solution to the equation 2(2-j)=4(J+10) ?<br><br> j =

sashaice [31]

$2(2-j)=4(j+10)\\
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8 0

3 years ago

Find the volume of a smaller wedge cut from a sphere of radius 66 by two planes that intersect along a diameter at an angle of π

djverab [1.8K]

Answer:

The answer is " $\frac{\pi a^3}{9}$ ".

Step-by-step explanation:

please find the complete question in the attached file.

Let the sphere center be (0,0,0) and then let the intersection diameter lie all along the z-axis.

So one of the collision plans is the xz-plane the other is the path via an xz-plane angle.

$\theta = \frac{\pi}{6}$

All appropriate region could then be indicated in spherical coordinates

$E= {(\rho, \theta, \phi) : 0 \geq \rho \geq a, 0 \geq \theta \geq \frac{\pi}{6}, 0 \geq \phi \geq \pi }$

Calculating the volume:

$\to v(E)=\int \int_{E} \int dV\\$

$=\int_{0}^{\frac{\pi}{6}} \int_{0}^{\pi} \int_{0}^{a} \rho^2 \sin \phi d \rho d \phi d \theta\\\\ =\int_{0}^{\frac{\pi}{6}} d \theta \int_{0}^{\pi} \sin \phi d \int_{0}^{a} \rho^2 d \rho\\\\= [\theta]^{\frac{\pi}{6}}_{0} [-\cos \phi]^{\pi}_{0} [\frac{\rho^3}{3}]^{a}_{0}\\\\= \frac{\pi}{6} [1+1] \frac{a^3}{3}\\\\=\frac{\pi a^3}{9}$

4 0

3 years ago