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3 years ago

6

Plss help I need this for my algebra class and have no idea what to do

1 answer:

Salsk061 [2.6K]3 years ago

7 0

Answer:

30t if t ≤ 2

25(t-2) + 60 if 2 < t ≤ 6

160 if 6 < t ≤ 10

Step-by-step explanation:

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kherson [118]

Answer:

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3 years ago

Convert the Cartesian equation (x 2 + y 2)2 = 4(x 2 - y 2) to a polar equation.

SashulF [63]

<u>ANSWER</u>

${r}^{2} = 4 \cos2\theta$

<u>EXPLANATION</u>

The Cartesian equation is

${( {x}^{2} + {y}^{2} )}^{2} = 4( {x}^{2} - {y}^{2} )$

We substitute

$x = r \cos( \theta)$

$y = r \sin( \theta)$

and

${x}^{2} + {y}^{2} = {r}^{2}$

This implies that

${( {r}^{2} )}^{2} = 4(( { r \cos\theta) }^{2} - {(r \sin\theta) }^{2} )$

Let us evaluate the exponents to get:

${r}^{4} = 4({ {r}^{2} \cos^{2}\theta } - {r}^{2} \sin^{2}\theta)$

Factor the RHS to get:

${r}^{4} = 4{r}^{2} ({ \cos^{2}\theta } - \sin^{2}\theta)$

Divide through by r²

${r}^{2} = 4 ({ \cos^{2}\theta } - \sin^{2}\theta)$

Apply the double angle identity

$\cos^{2}\theta -\sin^{2}\theta= \cos(2 \theta)$

The polar equation then becomes:

${r}^{2} = 4 \cos2\theta$

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3 years ago

If csc x = −5, find csc(−x) find the period and horizontal shift

ella [17]

Answer:

$\csc (-x) = - \csc x = - ( - 5) = 5$

Period is 2π

Horizontal shift = 0

Step-by-step explanation:

We know that $\sin (-x) = - \sin x$ and as $\csc x = \frac{1}{\sin x}$ , so, we can write $\csc (-x) = - \csc x$

Therefore, the if $\csc x = - 5$ then $\csc (-x) = - \csc x = - ( - 5) = 5$ (Answer)

Now, the period of $\csc x$ is 2π, hence the period of $\csc (- x)$ is also 2π. (Answer)

Again. $\csc ( - x) = \csc (-x + 0)$ , hence, the horizontal shift of $\csc (- x)$ is 0. (Answer)

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3 years ago

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VikaD [51]

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This means that the voter support for this candidate increased.

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3 years ago

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2 years ago