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Sedaia [141]
2 years ago
11

What is the fractional equivalent of the repeating decimal 03?

Mathematics
2 answers:
Degger [83]2 years ago
7 0
B. 3 10 is the correct answer
Dima020 [189]2 years ago
6 0
Ur answer is B 3 10 that is the correct answer!
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Solve for the variable X=5y for y
SSSSS [86.1K]

y= x/5. or x over 5. you are trying to isolate y. in other words you are trying to get y all by itself on one side of the equation. To do that, you need to divide 5 by both sides. It is important to do what you do on one side to the other because that keeps that values of the equation the same.
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4 years ago
The GCF of 148 is what ?
posledela
To find the gcf you need two numbers
8 0
3 years ago
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In which step did the veterinarian make the first error?<br> Step 1<br> Step 2<br> Step 3<br> Step 4
inna [77]

Answer:step 3 Step-by-step explanation: because it is right and i took the test edge 2021

4 0
3 years ago
A fruit punch recipe uses 3/4 cup of carbonated water for every 5 cups of fruit juice. Andy has only 4 cups of fruit juice. To m
Mama L [17]

Answer:

3/5 cup

Step-by-step explanation:

I start by changing the initial fraction of 3/4 into its decimal form of .75 so that it's easier to divide. I then divide it by 5 to find how much carbonated water you should use for a single cup of juice, which is .15 cups. Then, I multiply .15 times 4, which comes out to be .6, or 3/5.

8 0
3 years ago
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In how many different ways can a man divide 7 different gifts among his 3 children if the eldest is to receive 3 gifts and the o
krok68 [10]

The order in which gifts are received doesn't matter - if child X gets toy 1 then toy 2, it's the same as giving child X toy 2 then toy 1 - so we are counting combinations.

The eldest child receives 3 of the 7 gifts, so they have

\dbinom 73 = \dfrac{7!}{3!(7-3)!} = 35

possible choices of gifts.

The next child receives 2 of the remaining 4 gifts, so they have

\dbinom 42 = \dfrac{4!}{2!(4-2)!} = 6

choices.

The last child receives the remaining 2 gifts, and there is only

\dbinom22 = \dfrac{2!}{2!(2-2)!} = 1

way to select the gifts for them.

By the multiplication using, the total number of ways of distributing 7 gifts among 3 children in the prescribed way is 35 • 6 • 1 = 210.

7 0
2 years ago
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