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Mashutka [201]
3 years ago
14

QUICK QUICK ANSWER FIRST WITH AN EXPLANATION AND ANSWER YOU WILL GET THE CHANCE TO BE THE BRAINLIEST WITH AN BONUS/EXTRA POINTS

OF:20
The ratio of trees to flowering plants in Marsha’s backyard is shown on the ratio table. What number completes the ratio table to make an equivalent ratio? [Type your answer as a number.]

trees 2 12
plants 5 __?
Mathematics
1 answer:
Alex73 [517]3 years ago
6 0

Answer:

I pretty much think it is 30

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There were 30 students in the gym. 70% of the students in the gym were playing basketball. How many students in the gym were pla
seropon [69]

Answer:

21

Step-by-step explanation:

you multiply 30 by 0.7 and you get 21.

6 0
3 years ago
Read 2 more answers
If 2/3 ton of gravel covers 3/4 of a driveway, how many tons of gravel are required to cover the entire driveway? PLEASSE ANSWER
vivado [14]

If 2/3 ton of gravel covers 3/4 of a driveway, you need to divide 2/3 divided by 3 to find how much is needed for 1/4 of the driveway.

Steps:

1. 2/3 divided by 3 = 2/3 * 1/3. So 2/9.

2. Add 2/3 plus 2/9= 8/9 ton of gravel to cover the whole driveway.

Answer: 8/9

Hope this helped, and good luck with any future problems

7 0
3 years ago
Please solve the problem C.
anygoal [31]

Sure.

The common difference is 1/3

d = 8/15 - 1/5 = 8/15 - 3/15 = 5/15 = 1/3

d = 13/15 - 8/15 = 5/15 = 1/3

So the seventh term is

S_n = S_1 + (n-1)d

S_7 = (1/5)  + 6(1/3) = 11/5

The sum is the average of the first and last times the number of elements

(1/2)(1/5 + 11/5)(7) = 7(6/5) = 42/5 = 8.4

Answer: 42/5

3 0
3 years ago
Which quadrilateral and why?
Zigmanuir [339]
Personally you can’t measure you have to be in person but just double it on a piece of paper and you’ll find it out
6 0
3 years ago
If A ⊂ B, then A ∩ B = A ∪ B. always sometimes or never
Lisa [10]

Sometimes. Proof:

Assume A\subset B.

Let a\in A\cap B, so that a\in A and a\in B. Then of course a\in A\cup B, so A\cap B\subseteq A\cup B.

Now let a\in A\cup B, so that a\in A *or* a\in B. Since A\subset B, if a\in A then a\in B, so that a\in A\cap B. But if a\in B, then a\not\in A, so it's not always the case that a\in A\cap B, and therefore in general A\cup B\not\subseteq A\cap B.

6 0
3 years ago
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