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4 years ago

A 120 degree angle has its vertex at the center of a circle. What fraction of the circle does it cut out?

Mathematics

1 answer:

tino4ka555 [31]4 years ago

7 0

Answer:

Step-by-step explanation:

1/3.

A circle contains 360 degrees. 360 divided by 3 is 120.

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10 Pts + Brainliest for the best answer

LenaWriter [7]

m∠PQR = m∠BQR + m∠PQB

We have m∠PQR = 116° and m∠BQR = 75°. Substitute:

116° = 75° + m∠PQB <em>subtract 75° from both sides</em>

41° = m∠PQB

<h3>Answer: m∠PQB = 41°</h3>

3 0

3 years ago

Find three different ways to fill in operations in the lines below to make this equation true

MatroZZZ [7]

$1)\ \ \ 6-1+2-2=5+2-2=7-2=5\\\\2)\ \ \ (6-1)\cdot2:2=5\cdot2:2=10:2=5\\\\3)\ \ \ 6\cdot1-(2:2)=6-1=5$

3 0

4 years ago

you have housing and fixed expenses of $631.72/month and they are 33% of your realized income. What is our realized income per m

nikklg [1K]

631.72 = 33% of Realized Income
<span>631.72 / .33 = Realized Income </span>
<span>1914.30 = Realized Income</span>

6 0

3 years ago

Read 2 more answers

Sally and her friend Lucie started saving money for their vacation on the same day. Sally started with $150 and plans to add $5

solniwko [45]

Answer:

Their savings will be same after 25 weeks.

Step-by-step explanation:

The difference in both person's saving is $ 2/week

$5-$3=$2

The total difference in the starting point is $50 if we divide $50 by $2 we will get a number of 25 which is the number of weeks it will take both savings to reach at a same level

Calculation:

Starting point $150

Savings $5/week

5*25+150= $275

Starting point: $200

Savings 3$/week

3*25+200= $275

5 0

3 years ago

Let R be the region in the first quadrant bounded by the graphs of y =x^2 and y=2x, as shown in the figure above. The region R i

Kobotan [32]

Answer:

The area between the two functions is approximately 1.333 units.

Step-by-step explanation:

If I understand your question correctly, you're looking for the area surrounded by the the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

2 = x

So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

$a = \int\limits^2_0 {2x} \, dx - \int\limits^2_0 {x^2} \, dx\\\\a = x^2\left \{ {{x=2} \atop {x=0}} \right - \frac{x^3}{3}\left \{ {{x=2} \atop {x=0}} \right\\\\a = (2^2 - 0^2) - (\frac{2^3}{3} - \frac{0^3}{3})\\\\a = 2^2 - \frac{2^3}{3}\\a = 4 - 8/3\\a \approx 1.333$

So the correct answer is C, the area between the two functions is 4/3 units.

3 0

3 years ago