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kakasveta [241]
3 years ago
7

1A=bh

Mathematics
1 answer:
seropon [69]3 years ago
6 0

Answer:

120 yd2

Step-by-step explanation:

I assume you are solving for area.

The formula for a paralellogram (what it appears to be) is base times height.

The base in this equation is 10, and the height is 12. Multiplying this you get 120 yards squared.

For perimeter, you use pythagorean theorem to get the diagonal. 2\sqrt{61} . Then multiply 10 and 2\sqrt{61} by two, and then add them. 20+4\sqrt{61}.

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Fifty-eight boys were asked if they play football and/or baseball. Thirty-one of them said they do not play baseball. Sixteen of
iragen [17]

Answer:

5 play Football & Baseball; 22 don't play Football; 11 don't play baseball; 20 don't play either football or baseball. See captures attached. The missing one in your original answer + the one made in Excel.

Step-by-step explanation:

If you fill out the blanks the way explained in the answer, you can prove that the 3 sentences match perfectly. The sum of all of them totals 58.

1. Thirty-one of them said they do not play baseball.

2. Sixteen of them said they play football.

3. Twenty of them said they do not play baseball or football.

By the way, as I mentioned before, I also attached the missing image in your original question.

Have a great day!

7 0
3 years ago
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Scorpion4ik [409]

Answer:

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Step-by-step explanation:

8 0
3 years ago
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Factor completely<br> m^2 — Зm — 18
lilavasa [31]

Answer:

(m + 3) (m - 6)

Step-by-step explanation:

Simplify the following:

m^2 - 3 m - 18

The factors of -18 that sum to -3 are 3 and -6. So, m^2 - 3 m - 18 = (m + 3) (m - 6):

Answer: |

| (m + 3) (m - 6)

4 0
3 years ago
Let G be a connnectde graph with n vertices and m edges. supposed also that m = n. prove that G contains exactly one cycle
ElenaW [278]

Answer:

Contradiction

Step-by-step explanation:

Suppose that G has more than one cycle and let C be one of the cycles of G, if we remove one of the edges of C from G, then by our supposition the new graph G' would have a cycle. However, the number of edges of G' is equal to m-1=n-1 and G' has the same vertices of G, which means that n is the number of vertices of G. Therefore, the number of edges of G' is equal to the number of vertices of G' minus 1, which tells us that G' is a tree (it has no cycles), and so we get a contradiction.

7 0
3 years ago
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