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3 years ago

Which of the following describes how to translate the graph y = Ixl to obtain the graph of y = 1x - 11 - 1?

Mathematics

1 answer:

laiz [17]3 years ago

3 0

Answer:

Step-by-step explanation:

While your y = 1x - 11 - 1 is a function, it doesn't make sense in this context. I believe you meant y = |x - 11| - 1.

if that's correct, then the graph of this new function is obtained by translating the graph of y = |x| 11 units to the right and 1 unit down.

Unfortunately this result doesn't match any of the four answer choices. I ask that you go back to the original problem and type out the original of

y = 1x - 11 - 1.

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Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

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Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

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Step-by-step explanation:

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Which of the following is an extraneous solution of startroot 4 x 41 endroot = x 5? x = –8 x = –2 x = 2 x = 8

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The extraneous solution of startroot 4 x 41 endroot = x 5 will be A. -8.

<h3>What is an extraneous solution?</h3>

It should be noted that an extraneous solution simply means a root of a transformed equation which isn't part of the original equation.

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Learn more about extraneous solution on:

brainly.com/question/295656

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