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3 years ago

Simplify the square root of 50x^7y^10

1 answer:

7 0

$\sqrt{50x^{7}y^{10}} = \sqrt{50}* \sqrt{(x^{3})^{2}*x} * \sqrt{(y^{5})^{2}} =5 \sqrt{2}*x^{3} \sqrt{x} *y^{5}= \\ \\ 5x^{3}y^{5} \sqrt{2x}$

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12.) Write y = 1/6 x + 7 in standard form using integers.

Trava [24]

Option A

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<u>Solution:</u>

Given that we have to write $y = \frac{1}{6}x+7$ in standard form

The standard form of an equation is Ax + By = C

In this kind of equation, x and y are variables and A, B, and C are integers

Given equation is:

$y = \frac{1}{6}x+7$

Let us convert the above equation into standard form

$y = \frac{x}{6} + 7$

$y = \frac{x}{6} + \frac{7}{1}$

Make the denominator same in R.H.S

$y = \frac{x}{6} + \frac{7 \times 6}{1 \times 7}$

Solve the above equation

$y = \frac{x}{6} + \frac{42}{6}$

$y = \frac{x+42}{6}$

Move 6 from R.H.S to L.H.S

$6y = x + 42$

Bring all the terms to one side, leaving only constant on R.H.S

$-x + 6y = 42$

The above equation is of form Ax + By = C

Thus option A is correct

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