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Studentka2010 [4]
2 years ago
14

Eliminate the parameter. x = 3 cos t, y = 3 sin t

Mathematics
1 answer:
ElenaW [278]2 years ago
4 0

Answer:

x^2+y^2 = 3^2

Step-by-step explanation:

We need to eliminate the parameter t

Given:

x = 3 cos t

y = 3 sin t

Squaring the above both equations

(x)^2=(3 cos t)^2

(y)^2 =(3 sin t)^2

x^2 = 3^2 cos^2t

y^2=3^2 sin^2t

Now adding both equations

x^2+y^2=3^2 cos^2t+3^2 sin^2t

Taking 3^2 common

x^2+y^2=3^2 (cos^2t+sin^2t)

We know that cos^2t+sin^2t = 1

so, putting the value

x^2+y^2=3^2(1)

x^2+y^2 = 3^2

Hence the parameter t is eliminated.

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The graph shows the function f(x).
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The equation that represents the given data is  \RM \\f(x) = -\sqrt[3]{-x} , Option D is the correct answer.

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It is given that

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2 years ago
-4/5(-8x+6)=1+7 Please Help
bagirrra123 [75]

Answer:

x = 2

Step-by-step explanation:

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PLUG IN

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