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-BARSIC- [3]
3 years ago
10

(5m-2n) (25m^2 + 10mn + 4n^2)

Mathematics
1 answer:
RideAnS [48]3 years ago
4 0

Answer:

<h2><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em></h2>

<em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em><em><u> </u></em>

(5m - 2n)(25 {m}^{2} + 10mn + 4 {n}^{2})

5m(2 {m}^{2}  + 10mn + 4 {n}^{2} ) - 2n(25 {m}^{2} + 10mn + 4 {n}^{2} )

10 {m}^{3} + 50 {m}^{2}n + 20m {n}^{2}  - \\ 50 {m}^{2} n - 20m {n}^{2} - 8 {n}^{3}

10 {m}^{3} - 8 {n}^{3} (ans)

<h3><em><u>H</u></em><em><u>O</u></em><em><u>P</u></em><em><u>E</u></em><em><u> </u></em><em><u>I</u></em><em><u>T</u></em><em><u> </u></em><em><u>H</u></em><em><u>E</u></em><em><u>L</u></em><em><u>P</u></em><em><u>S</u></em><em><u> </u></em></h3>

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This question s incomplete, the complete question is;

The Watson family and the Thompson family each used their sprinklers last summer. The Watson family's sprinkler was used for 15 hours. The Thompson family's sprinkler was used for 30 hours.

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Answer:

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Step-by-step explanation:

Given the data in the question;

let water p rate for Watson family and the Thompson family sprinklers be represented by x and y respectively

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x + y = 55  ----------------equ1

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also

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(55 - y) + 2y = 70

- y + 2y = 70 - 55

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Therefore, The Watson family sprinkler is 40 L/hr while Thompson family sprinkler is 15 L/hr

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If a dog has 2,000,000 toys and he gives 900,000 away. Then gets 2,000 more, also looses 2,000,000. He's sad but then also got 5
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See below.

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He does not have enough to loose 2,000,000 at that point, so this whole problem is nonsense.

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