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Makovka662 [10]
3 years ago
5

One endpoint (-1,-5) Midpoint (2,3)

Mathematics
1 answer:
Alecsey [184]3 years ago
4 0
I don’t know the answer but I need a answer to this
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Please help the following questions
insens350 [35]

Answer:

<em>Option D</em>

Step-by-step explanation:

Assume that these polygons were similar. You could tell that if they were, ABCD ~ EFGH, and is already noted by the answer choices. We can conclude that AB ~ EF, CD ~ GH and so on. To prove that these shapes are similar, we must form a like proportionality among these side;

EF / AB = GH / CD\\5 / 8 = 4 / 2.5,\\0.625 \neq 1.6\\\\Conclusion - Option D

As the proportions we formed were to equal to one another,<em> the solution must be option D!</em>

6 0
3 years ago
What is it called when your equation is 0x=5
laila [671]
I just write not possible
8 0
3 years ago
What is 4 to the 7th power ?
bonufazy [111]
4^7=16,384 i  used a calulator

5 0
3 years ago
Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r
erastova [34]

Answer:

The probability is \frac{56!}{64!}

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 64 \choose 8 .

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function  f : A \rightarrow A , with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}

7 0
3 years ago
What equation represents this graph?
Alexxandr [17]
It should be d. y=2/3x+2
7 0
2 years ago
Read 2 more answers
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