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natali 33 [55]
3 years ago
9

Simplify each expression as much as possible, and rationalize denominators when applicable. √18x/(3√x^5)=?

Mathematics
1 answer:
Anni [7]3 years ago
8 0

Answer:

  (√2)/x²

Step-by-step explanation:

\dfrac{\sqrt{18x}}{3\sqrt{x^5}}=\sqrt{\dfrac{18x}{9x^5}}=\sqrt{\dfrac{2}{x^4}}=\dfrac{\sqrt{2}}{x^2}

_____

Or, maybe you mean ...

\dfrac{\sqrt{18}x}{3\sqrt{x^5}}=\sqrt{\dfrac{18x^2}{9x^5}}=\sqrt{\dfrac{2x}{x^4}}=\dfrac{\sqrt{2x}}{x^2}

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3 0
3 years ago
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mrs_skeptik [129]

Given:

Line a passes through (2, 10) and (4, 13).

Line b passes through (4, 9) and (6, 12).

Line c passes through (2, 10) and (4, 9).

To find:

Which of the lines, if any are perpendicular.

Solution:

If a line passes through two points, then the slope of line is

m=\dfrac{y_2-y_1}{x_2-x_1}

Line a passes through (2, 10) and (4, 13).  So, slope of this line is

m_a=\dfrac{13-10}{4-2}=\dfrac{3}{2}

Line b passes through (4, 9) and (6, 12).  So, slope of this line is

m_b=\dfrac{12-9}{6-4}=\dfrac{3}{2}

Line c passes through (2, 10) and (4,9).  So, slope of this line is

m_c=\dfrac{9-10}{4-2}=\dfrac{-1}{2}

Product of slopes of to perpendicular lines is -1.

m_a\cdot m_b=\dfrac{3}{2}\times \dfrac{3}{2}=\dfrac{9}{4}\neq -1

m_b\cdot m_c=\dfrac{3}{2}\times \dfrac{-1}{2}=\dfrac{-3}{4}\neq -1

m_a\cdot m_c=\dfrac{3}{2}\times \dfrac{-1}{2}=\dfrac{-3}{4}\neq -1

Therefore, any of these lines are not perpendicular to each other.

8 0
3 years ago
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solmaris [256]

Answer:

add them both. 4,3 is your answer

Step-by-step explanation:

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