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3 years ago

Simplify. Assume x and y are nonnegative.

Mathematics

2 answers:

AnnZ [28]3 years ago

7 0

$9\sqrt{56x^7y^{12}}=\\\\9\sqrt{4x^6\cdot14x \cdot (y^6)^2}=\\\\9y^6\sqrt{14x\cdot (2x^3)^2}=\\\\9y^6\cdot 2x^3\sqrt{14x}=\\\\18x^3y^6\sqrt{14x}$

Taya2010 [7]3 years ago

3 0

Here is the answer :)

I actually answered first, but someone else copied me and reported me unfairly. I’ve typed in the step by steps.

Hope this helps

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O is the center of the regular octagon below. Find its area. Round to the nearest tenth

Sedbober [7]

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Step-by-step explanation: to much work I thought to hard that’s the answer your welcome buddy

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2 years ago

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3 years ago

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skad [1K]

Since the lines have the same slopes, hence they are parallel lines

<h3>How to determine the relationship between lines</h3>

We can determine the relations by knowing the slopes of the line

For the line with coordinates (9, –5) and (5, –2)

Slope = -2+5/5-9
Slope = -3/4

For the line with coordinates (–4, –2) and (–8, 1)

Slope = 1+2/-8+4
Slope = -3/4

Since the lines have the same slopes, hence they are parallel lines

Learn more on parallel lines here: brainly.com/question/16742265

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2 years ago

Here are the endpoints of the segments BC, FG, and JK.<br> B, −67

yulyashka [42]

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{[-4 - (-6)]^2 + [4 - 7]^2}\implies BC=\sqrt{(-4+6)^2+(-3)^2} \\\\\\ BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}$

$F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}$

$JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill$

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2 years ago