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erastova [34]
3 years ago
7

WILL GIVE A BRAINLEST

Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
5 0
Two end points
One direction

Hope this helps xx
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Find the area of sector AOB. Express your answer in terms of pi and rounded to the nearest tenth.
STatiana [176]
63\pi

The sector is \frac{70}{360}; there are 360 degrees in a circle.
You use the formula \pi r^{2}.
18cm is the radius, and then you square it to get 324.
Multiply 324 by \frac{70}{360}
You get 63\pi
7 0
3 years ago
Juanita is having a party at her house she is inviting 12 people.there will be 3 times as many boys as girls invited. let x repr
Margaret [11]
X + 3x = 12 

(Or 4x=12)

the first x is the number of girls then the second x is the girls times 3, which is how many boys there are.
8 0
3 years ago
THIS IS MULTIPLE CHOICE!
Troyanec [42]

Answer:

there is a cluster from 1-4

there is a gap from 5-7

The spread is from 1-8

Step-by-step explanation:

These are all true if you look... there is a bunch from 1-4, there is none from 5-7, and all of them are within 1-8, but  the highest amount isn't at 3.

7 0
3 years ago
Lola and Brandon had a running race. They ran 3/10 of a mile. Lola was in the lead for 4/5 of the distance. For what fraction of
Kruka [31]

Answer:

Lola was in lead for \frac{6}{25} of a mile.

Step-by-step explanation:

We have been given that Lola and Brandon had a running race. They ran 3/10 of a mile. Lola was in the lead for 4/5 of the distance.

To find the fraction of a mile for which Lola was in the lead, we need to find 4/5 of 3/10 as:

\text{Fraction for which Lola was in lead}=\frac{3}{10}\times\frac{4}{5}

\text{Fraction for which Lola was in lead}=\frac{3}{5}\times\frac{2}{5}

\text{Fraction for which Lola was in lead}=\frac{3\times2}{5\times5}

\text{Fraction for which Lola was in lead}=\frac{6}{25}

Therefore, Lola was in lead for \frac{6}{25} of a mile.

7 0
3 years ago
10 minus three times a number is no more than 61​
Elena L [17]

Answer:

n ≥ -17

Step-by-step explanation:

Writing a symbolic inequality, we get:

10 - 3n ≤ 61

Solve this for 3n by adding 3n to both sides of this equation:

10 ≤ 61 + 3n

Solve for n by subtracting 61 from both sides and then dividing all of the resulting terms by 3:

-51 ≤ 3n (divide both sides by 3):

-17 ≤ n, or

n ≥ -17

8 0
3 years ago
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