<img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B11%7D%20%3D%5Cfrac%7Br%7D%7B13%7D%3F" id="TexFormula1" title="\frac{2}{11} =\

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Slav-nsk [51]

2 years ago

frac{r}{13}?" alt="\frac{2}{11} =\frac{r}{13}?" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

notka56 [123]2 years ago

4 0

I hope that it helps you..

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This makes no sense can someone help me please

Verizon [17]

It just has to equal 90° so that they are parallel so 16x-6=90
16x=96 (add 6 to both sides)
x=6(divide by 16)

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3 years ago

A Salt Marsh Harvest Mouse ran a 360 centimeter straight course race in 9 seconds. How fast did it run?

Gnoma [55]

Answer: 40 Centimeters per second

Step-by-step explanation:

360/9 = 40

5 0

3 years ago

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2(x-4)+?=8x-9 what is the missing polynomial

Alex777 [14]

First use distributive property on the left side
2x - 8 + _________ = 8x - 9
What do u need to make 2x - 8 into 8x - 9
6x - 1

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3 years ago

Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess

Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2$