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Wittaler [7]
3 years ago
9

It’s confusing, would the answer be square root 7 over 4?

Mathematics
1 answer:
Colt1911 [192]3 years ago
4 0

Answer:

\tan U = 1

Step-by-step explanation:

\text{Apply Pythagorean theorem,}\\\\~~~~~~\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2\\\\\implies UT^2 = UV^2 +VT^2\\\\\implies UV^2 = UT^2 -VT^2\\\\\implies UV^2 = \left(7\sqrt 2 \right)^2 - 7^2\\\\\implies UV^2 = 49(2)-49\\\\\implies UV^2 = 49\\\\\implies UV = \sqrt{49} \\\\ \implies UV = 7

\text{Now,}\\\\~~~~~~~~\tan \theta = \dfrac{\text{Perpendicular}}{\text{Base}}\\\\\\\implies \tan U = \dfrac{7}{7}\\\\\\\implies \tan U = 1

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What is an equation of the line perpendicular to y=-x-2 and through (-2, 4)?
Salsk061 [2.6K]

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\mathrm{Find\:the\:line\:}\mathbf{y=mx+b}\mathrm{\:perpendicular\:to\:}y=-x-2\mathrm{\:that\:passes\:through\:}\left(-2,\:4\right)

\mathrm{For\:a\:line\:equation\:for\:the\:form\:of\:}\mathbf{y=mx+b}\mathrm{,\:the\:slope\:is\:}\mathbf{m}

m=-1

\mathrm{The\:perpendicular\:slope\:is\:the\:negative\:reciprocal\:of\:the\:given\:slope}

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m_p=1

\mathrm{Compute\:the\:line\:equation\:}\mathbf{y=mx+b}\mathrm{\:for\:slope\:m=}1\mathrm{\:and\:passing\:through\:}\left(-2,\:4\right)\mathrm{Plug\:the\:slope\:}1\mathrm{\:into\:}y=mx+b

y=x+b

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\mathrm{Add\:}2\mathrm{\:to\:both\:sides}

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\text{Simplify}

b=6

\mathrm{Construct\:the\:line\:equation\:}\mathbf{y=mx+b}\mathrm{\:where\:}\mathbf{m}=1\mathrm{\:and\:}\mathbf{b}=6

y=x+6

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