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azamat
4 years ago
5

Please answer this question!!thanks xx

Mathematics
1 answer:
Simora [160]4 years ago
7 0
180-(180-49-48)
180-(83)
97

180-((180-141)+(180-76))
180-(39+104)
180-143
137

180-((180-101)+60)
180-(79+60)
180-139
41

360-147-79
213-79
139

180-(360-109*2)
180-(360-218)
180-142
38

(180-118)/2
62/2
31

Hope this helps :)
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Ivana printer prints 40 pages in 5 minutes.How many pages will the printer print in 8 minutes
Lana71 [14]

64 pages

divide 40 by 5 for pages printed in 1 minute then multiply this by 8

\frac{40}{5} × 8 = 8 × 8 = 64 pages


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4 years ago
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Point (, ) is dilated from the origin by scale factor = . What are the coordinates of point ′?
Igoryamba

Answer:idk how to do this im in 7th grade

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The ratio of the lengths of the corresponding sides of two rectangles is 8:3. The area of the larger rectangle is 320 ft2. What
cluponka [151]
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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
Kat is painting the edge of a triangular stage prop with reflective orange paint. The lengths of the edges of the triangle are (
Eva8 [605]
Plug 4 in for every x

(3*4)-4=8
(4^2)-1=15 [(4^2) is 4*4]
(2*(4^2))-15= 15 [because of PEMDAS the exponent comes first and then multiply by 2 then subtract]

Perimeter is the sum of the lengths so add up all the sides from above
8+15+15=38
3 0
3 years ago
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