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kicyunya [14]
2 years ago
12

PLEASE ANSWER HONESTLY (PLEASE BE HELPFULL I HAVE BEEN STUCK ON THIS PROBLEM) WILL MARK BRANIEST IF CORRECT. (no links my comput

er is slow and no ctrl c ctrol v if possible (see picture below)

Mathematics
2 answers:
PilotLPTM [1.2K]2 years ago
4 0
The answer is B- 36pi. I just divided 108pi by 3. Hope this helps! If you’re confused, feel free to ask me to explain.

:)
Svetach [21]2 years ago
3 0

Answer:

If im correct it's 36 pie.

Step-by-step explanation:

I believe this as 1/3 of 108 is 36.

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Tom went to the supermarket and could not spend more than $30.00. He spent $9 on dairy
valina [46]

Answer:

it will be c

Step-by-step explanation:

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3 years ago
Lola walks 4/10 mile to her friends house. then she walks 5/10 mile to the store. how far does she wall in all?
Masteriza [31]
We know that <span>Lola walks 4/10 mile to her friends house. then she walks 5/10 mile to the store. 

To solve the problem we have to use addition. 

4/10 + 5/10 = 9/10 

Lola walks 9/10 miles. </span>
8 0
3 years ago
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Matt's phone bill consist of an initial monthly charge of $30 and then he pays $4.50 per gigabyte of internet he uses. If he spe
ivanzaharov [21]

Answer:

5 gigabytes

Step-by-step explanation:

30+4.50x=52.50 (equation)

-30           -30

4.50x=52.50

/4.50   /4.50

x= 5

5 gigabytes

5 0
3 years ago
Some of the students three scores is 231. If the first is 20 points more than the second, and the sum of the first two is 6 more
Aleks [24]

Answer:

The first score is 109

Step-by-step explanation:

I am assuming that in the first sentence of the question, you meant:

Sum of the students three scores is 231...

First, let the scores of the first second and third student be a, b and c respectively. We are told that:

a + b + c = 231 . . . . . . . .(1)         (sum of students three scores is 231)

a = b + 20 . . . . . . . . . . . (2)        (the first is 20 points more than the second)

a + b = 6c . . . . . . . . . . . .(3)        (sum of the first two is 6 more times the third)

required, find a.

substituting the value of (a + b) in equation (3) into equation (1), we will have the following:

since a + b = 6c . . . (3)

a + b + c = 231 . . . . . (1), becomes,

(a + b) + c = 231

(6c)  + c = 231

7c = 231 (divide both sides by 7)

c = 231 ÷ 7 = 33

∴ c = 33

Next, from equation (2), we know that a = b + 20; this can also be written as:

a - 20 = b

∴ b = a - 20 . . . . . . . (4)

Finally, putting the value of b in equation (4) and the value of c calculated above into equation 1, ( a + b + c = 231), we have the following:

a + (a - 20) + 33 = 231

(a + a) - 20 + 33 = 231

2a + 13 = 231

2a = 231 - 13 = 218

a = 218 ÷ 2 = 109

∴ a = 109

we can also calculate for 'b' by substituting for the value of 'a' in equation 4

b = a - 20 = 109 - 20 = 89.

and to test if the values of a, b and c are correct:

a + b + c = 231

109 + 89 + 33 = 231

4 0
3 years ago
A and B are two events.
Brut [27]
I think that the answer is A.
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3 years ago
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