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3 years ago

Can someone please help and explain this to what the answer is I’ll give brainliest

Mathematics

1 answer:

CaHeK987 [17]3 years ago

4 0

Answer:

BOTH THE QUESTIONS ARE SAME!!

When finding a perpendicular slope, you need to <u>change the signs</u> and <u>flip the reciprocal</u>

The answer to your question is:

m = -2

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3 years ago

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You are given the polar curve r = e^θ

galina1969 [7]

The tangent to $r(\theta)=e^\theta$ has slope $\frac{\mathrm dy}{\mathrm dx}$ , where

$\begin{cases}x(\theta)=r(\theta)\cos\theta\\y(\theta)=r(\theta)=\sin\theta\end{cases}$

By the chain rule, we have

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

and by the product rule,

$\dfrac{\mathrm dx}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\cos\theta-r(\theta)\sin\theta$

$\dfrac{\mathrm dy}{\mathrm d\theta}=\dfrac{\mathrm dr}{\mathrm d\theta}\sin\theta+r(\theta)\cos\theta$

so that with $\frac{\mathrm dr}{\mathrm d\theta}=e^\theta$ , we get

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^\theta\sin\theta+e^\theta\cos\theta}{e^\theta\cos\theta-e^\theta\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}=-\dfrac{1+\sin(2\theta)}{\cos(2\theta)}$

The tangent line is horizontal when the slope is 0; this happens for

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i.e at the points

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and

$(r,\theta)=\left(e^{7\pi/4},\dfrac{7\pi}4\right)$

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3 years ago

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8_murik_8 [283]

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3 years ago

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Answer:

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Given expressions are $2x^2y^2,-3xy^3,-4x^3y^2,x^3,8.4x^4y^2,5x^4y,-xy^3$

<h3>To find the first term of the expression below to create a polynomial with a degree of 5 written in standard form :</h3>

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liubo4ka [24]

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