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Hatshy [7]
3 years ago
12

Complete the long division problem on your own sheet of paper. Then, fill in the digits.

Mathematics
1 answer:
Bas_tet [7]3 years ago
7 0

Answer:

1864 R 2

Step-by-step explanation:

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Help please solve for the given variable<br> -8 (7k - 5) + 7(7k - 7) = 1 - 5k - k - 7
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- 8(7k-5) + 7(7k-7) = 1 - 5k - k - 7
- 56k + 40 + 49k - 49 = 1 - 6k - 7
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8 0
4 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

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3 years ago
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Vanyuwa [196]

Answer:

19750

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Slope of 2/3 and x-intercept of -3
Vilka [71]

Answer:

\frac{2}{3},-3

Step-by-step explanation:

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