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2 years ago

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The coordinates of the vertices of the preimage of a triangle are (2, 1), (3, 4), and (4, 1). The coordinates of the vertices of

the image are (3, 3), (4, 6), and (5, 3). How far and in what direction was the triangle translated?

1 answer:

lbvjy [14]2 years ago

6 0

Translated 1 unit up and 2 units to the right

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Evaluate the expression<br><br> x^2 (y - 2)<br> —————, if x=3, and y= -1<br> 3y

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Answer:

$\huge\boxed{\sf 9}$

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$\displaystyle = \frac{x^2(y-2)}{3y} \\\\Put \ x = 3, \ y = -1\\\\= \frac{(3)^2(-1-2)}{3(-1)}\\\\= \frac{9(-3)}{-3} \\\\= 9 \\\\ \rule[225]{225}{2}$

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2 years ago

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Y = –3x – 5 help I don’t know the answer please

earnstyle [38]

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2 years ago

Is sin theta=5/6, what are the values of cos theta and tan theta?

mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}$

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2 years ago