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Lerok [7]
3 years ago
5

1.What are the values of the 6s in 36,621?

Mathematics
1 answer:
enot [183]3 years ago
8 0
1) The value of 6s in 36,621 are 6,000 and 600.

first 6 is in the thousands place value while the second 6 is in the hundreds place value.
6 x 1000 = 6,000 and 6 x 100 = 600

2) The first 4 is ten times greater than the second 4 is shown in 354,426

the first 4 is in the thousands place value while the second 4 is in the hundreds place value
4 x 1000 = 4000 and 4 x 100 = 400

based on the question the first 4, which is 4000, is ten times greater than the second 4, which is 400.
we can check if this statement is true either by dividing 4000 by ten or multiplying 400 by ten.

4000/10 = 400 ; 400 x 10 = 4000. therefor, out of the 4 numbers given, the correct one is 354,426

3) The relationship of the two 5s in 552 is shown in 500 ÷ 50 = 10

based on its place value, the first five is in the hundreds place and the second five is in the tens place.

500 ÷ 50 = 10. This equation states that 500 is ten times greater than 50.

4) In 2,077 the first 7 is greater than the second 7 by 10 times.

using place value, first 7 is in the tens place while second 7 is in the ones place.

first 7 is 70; second 7 is 7. we divide 70 by 7 to know how many number of times 70 is greater than 7.

70 ÷ 7 = 10 times.


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Answer:

\dfrac{(a+b)}{ab}

Step-by-step explanation:

The given expression is :

\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}

It can be solved as follows :

\dfrac{a}{ab-b^2}+\dfrac{b}{ab-a^2}=\dfrac{a}{b(a-b)}+\dfrac{b}{a(b-a)}\\\\=\dfrac{a}{b(a-b)}+\dfrac{b}{-a(-b+a)}\\\\=\dfrac{1}{a-b}(\dfrac{a}{b}-\dfrac{b}{a})\\\\=\dfrac{a^2-b^2}{ab(a-b)}\\\\=\dfrac{(a-b)(a+b)}{ab(a-b)}\\\\=\dfrac{(a+b)}{ab}

So, the solution of the given expression is equal to \dfrac{(a+b)}{ab}.

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3 years ago
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Answer:

x=4.875

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Step-by-step explanation:

4x – 2y = 10

2x-y=10

y=2x-10

4x + 6y = 18

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x=4.875

y=2(4.875)-10

y=-0.25

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Sonbull [250]

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