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3 years ago

14

True of false !!! The triangle shown below must be congruent

2 answers:

user100 [1]3 years ago

6 0

False.

Right triangles do not have to be congruent.

:)

slava [35]3 years ago

5 0

It is false please mark brainliest

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2 years ago

Find all points on the curve x=4cos(t),y=4sin(t) that have the slope of 12.

Marianna [84]

Answer:

$\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$ .

Step-by-step explanation:

We need to find all the points on the curve x=4cos(t),y=4sin(t) that have the slope of 1/2.

$x=4cos (t)$

$\dfrac{dx}{dt}=-4sin (t)$

$y=4sin (t)$

$\dfrac{dy}{dt}=4cos (t)$

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$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$

$\dfrac{dy}{dx}=4cos (t)\times \dfrac{1}{-4sin (t)}$

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So, slope of the curve is $-\cot t$ .

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$-\tan t=2$

$\tan t=-2$ ...(1)

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$\sec^2t=1+4$

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$\sin t=\pm \sqrt{1-\dfrac{1}{5}}$

$\sin t=\pm \sqrt{\dfrac{4}{5}}$

$\sin t=\pm \dfrac{2}{\sqrt{5}}$

It equation (1), tan(t) is negative. So, sin and cos have different signs.

If $\sin t= \dfrac{2}{\sqrt{5}}$ , then $\cos t=- \dfrac{1}{\sqrt{5}}$ .

$x=4cos (t)=-\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=\dfrac{8}{\sqrt{5}}$

If $\sin t=- \dfrac{2}{\sqrt{5}}$ , then $\cos t= \dfrac{1}{\sqrt{5}}$ .

$x=4cos (t)=\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=-\dfrac{8}{\sqrt{5}}$

Therefore, the two points are $\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$ .

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