All categories

2 years ago

Arc CD is One-fourth of the circumference of a circle. What is the radian measure of the central angle?

Mathematics

1 answer:

Alex2 years ago

8 0

Answer:

$\frac{\pi }{2}$

Step-by-step explanation:

Since the arc is $\frac{1}{4}$ of the circumference, then

central angle = $\frac{1}{4}$ × 2π ← angle in whole circle

central angle = $\frac{2\pi }{4}$ = $\frac{\pi }{2}$

You might be interested in

A research submarine descends 65 meters per minutes to a depth of 4485 m, conducts experiments, and then ascends at a rate of 65

sukhopar [10]

Answer:

Step-by-step explanation:

speed =65m/mins = 65/60 = 1.083m/s

Time taken to go down,

t = 4485/1.083

= 4141.37 s

Total time gotten = 12 hrs = 12*60*60

= 43200s

Amount of time left to conduct experiment

= 43200-4141.37

= 39058.63s

= 39058.63/3600

= 10.85 hrs

7 0

1 year ago

Read 2 more answers

You work two jobs. You earn $11.90 per hour as a salesperson and $10.50 per hour stocking shelves. Your combined earnings this m

Delicious77 [7]

178.50 = 11.90y + 10.50x ⇒ equation in standard form.

4 0

2 years ago

Select all the pairs that could be reasonable approximations for the diameter and circumference of a cirlce 5 meters and 22 mete

bazaltina [42]

Answer:

Option a) circle 5 meters and 22 meters

Step-by-step explanation:

We are given the following information in the question:

A pair of diameter and the circumference is given. We have to find a correct approximations for the diameter and circumference.

a) circle 5 meters and 22 meters

$\text{Diameter} = 5\text{ meters}\\\text{Circumference} = \pi d = 3.14\times 5 = 15.7\text{ meters}$

b) 19 inches and 50 inches

$\text{Diameter} = 19\text{ inches}\\\text{Circumference} = \pi d = 3.14\times 19 = 59.66\text{ inches}$

c) 33 centimeters and 80 centimeters

$\text{Diameter} = 33\text{ centimeters}\\\text{Circumference} = \pi d = 3.14\times 33 =103.62\text{ centimeters}$

Thus, no pair gives a reasonable approximation. Only the circle with diameter 5 and circumference 22 meters have closest approximation.

6 0

2 years ago

Which expression has the same GCF as 15x2 - 21x?

Mice21 [21]

Answer:

$15x^2 +21 x$

Step-by-step explanation:

We have the following expression:

$15x^2 -21 x$

We can find the GCF with the following steps.

1. Find the GCF for the numerical part 15 and -21

The factors for 15 are : 1,3,5,15

The factors for -21 are -21,-7,-3,-1,1,3,7,21

And the common factors are 1,3

The GCF for the numerical part is 3*1 = 3

2. Find the GCF for the variable

The factors of x^2 are : x*x

The factors of x are: x

The common factors are x for this case, so then the GCF for the variable part is x

3. Multiply the two results from steps 1 and 2

GCf= 3*x = 3x

If we consider the new expression:

$15x^2 +21x$

We can find the GCF with the following steps.

1. Find the GCF for the numerical part 15 and -21

The factors for 15 are : 1,3,5,15

The factors for -21 are 1,3,7,21

And the common factors are 1,3

The GCF for the numerical part is 3*1 = 3

2. Find the GCF for the variable

The factors of x^2 are : x*x

The factors of x are: x

The common factors are x for this case, so then the GCF for the variable part is x

3. Multiply the two results from steps 1 and 2

GCf= 3*x = 3x

And as we can see both GCF are equal. So then we satisfy the condition required.

5 0

1 year ago

Read 2 more answers

A turtle and a rabbit engage in a footrace over a distance of 4.00 km. The rabbit runs 0.500 km and then stops for a 90.0-min na

Oxana [17]

Answer:

(a) 2.29 km/h

(b) 9 km/h

Step-by-step explanation:

For part (a) you have to apply<em> the average speed formula</em>, which is defined by:

$v=\frac{d}{t}$

where d is the total distance traveled and t is the total time needed.

$d=\frac{4.00}{1.75}=2.29$ km/h

For part (b) you have to calculate the running time (T) , which is the total time of the race minus the nap time:

The nap time in hours is:

90/60 = 1.5 h (because there are 60 minutes in one hour)

The running time is:

T= 1.75 - 1.5 = 0.25 h

Let t1 represent the time before the nap and t2 the time after the nap:

t1+t2 = T

t1+t2 = 0.25

You have to apply the formula d=vt before and after the nap:

-Before the nap, the distance traveled was 0.50 km

0.50 = v1t1

-Afer the nap, the distance traveled was 3.50 km

3.50=v2t2

But v2=2v1 (because after the nap the rabbit runs twice as fast)

You have to solve the system of equations:

t1=0.25-t2 (I)

v1t1=0.50 (II)

2v1t2=3.50 (III)

Replacing (I) in (II)

v1(0.25-t2)=0.50

Applying distributive property and solving:

0.25v1-v1t2=.050

For (III) you have that v1t2=3.50/2=1.75. Hence:

0.25v1-1.75=0.50

Solving for v1:

v1 = 9 km/h

7 0

1 year ago