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2 years ago

The radius of a circle is 6 inches. What is the length of a 135-degree arc?

1 answer:

8 0

14.14 inches

Convert 135° into radians. 135*2pi/360=3pi/4.
6*3pi/4=18pi/4=9pi/2=14.14

Another viable method is to do 2pi*6*135/360=14.14

You might be interested in

Evaluate 4(6) + 7 (5 - 3 ) A: 118 B: 80 C: 10 D: 38

ivanzaharov [21]

4(6) + 7 (5 - 3 )
= 24 + 7(2)
= 24 + 14
= 38

answer
D: 38

5 0

2 years ago

Read 2 more answers

150 - 1/2s = 120 + 1/4s

agasfer [191]

Answer:

s=40

Step-by-step explanation:

Combine s and 1/2.

150−s/2=120+1/4s

Combine 1/4and s.

150−s/2=120+s/4

Move all terms containing s to the left side of the equation.

150−3s/4=120

Move all terms not containing s to the right side of the equation.

−3s/4=−30

Multiply both sides of the equation by −4/3.

−4/3 x (−3s/4)=−4/3 x−30 => S=40

3 0

3 years ago

Read 2 more answers

What is the value of x

Nesterboy [21]

Answer:

$x = 2$

Step-by-step explanation:

First simplify the equation.

$\frac{7}{9}x -\frac{4}{9} =\frac{1}{4} +\frac{5}{12}$
$\frac{3}{9}x= \frac{3}{12} +\frac{5}{12}$
$\frac{3}{9}x=\frac{8}{12}$
$\frac{1}{3}x=\frac{2}{3}$

Now multiply both sides by 3.

$x = 2$

I hope it helps you. Please choose my answer as the BRAINLIEST.

7 0

2 years ago

Read 2 more answers

Find all values of x in the interval [0, 2π] that satisfy the equation. (enter your answers as a comma-separated list.) 6 sin(2x

anygoal [31]

6 sin 2x = 6 cos x Using the identity sin 2x = 2 sin x cos x:-

12 sin x cos x = 6 cos x

6 cos x ( 2 sin x - 1) = 0

either 6 cos x = 0 or 2 sin x - 1 = 0 so sin x = 1/2

x = pi/2, 3pi/2 , pi/6, 5pi / 6

answer is ( pi/6, pi/2, 5pi/6, 2pi/2)

6 0

3 years ago

The distance d of a particle moving in a straight line is given by d(t) = 2t3 + 5t – 2, where t is given in seconds and d is mea

Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

Given the distance, d(t) of a particle moving in a straight line at any time t is:

$d(t) = 2t^3 + 5t - 2, $ where t is given in seconds and d is measured in meters.$

To find an expression for the instantaneous velocity v(t) of the particle at any given point in time, we take the derivative of d(t).

$v(t)=\dfrac{d}{dt}\\\\v(t) =\dfrac{d}{dt}(2t^3 + 5t - 2) =3(2)t^{3-1}+5t^{1-1}\\\\v(t)=6t^2+5$

The correct option is C.

6 0

3 years ago