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andreyandreev [35.5K]
3 years ago
11

(x-h)^2+(y-k)^2=r^2

Mathematics
1 answer:
dimaraw [331]3 years ago
7 0

Answer:

(x-2)^2 + (y - 5)^2 = 16

Step-by-step explanation:

Given the standard form of an equation of a circle

(x-h)^2+(y-k)^2=r^2

If (h, k) = (5,2) and r = 4

Substitute

(x-2)^2 + (y - 5)^2 = 4^2

(x-2)^2 + (y - 5)^2 = 16

Hence the required equation is (x-2)^2 + (y - 5)^2 = 16

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

5 0
3 years ago
How do you solve 0.8=6-2.5x
photoshop1234 [79]
X=2.08

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Read 2 more answers
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Answer:

option 2

Step-by-step explanation:

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