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3 years ago

State the range of the relation

Mathematics

1 answer:

Finger [1]3 years ago

3 0

The range of the given relation is D. R = {-1, 3, 5, 8}.

Step-by-step explanation:

Step 1:

The range of a relation is the second set of values while the domain constitutes the first set of values.

There are 4 given relations with two sets of values so there would be 4 domain values and 4 range values.

Step 2:

The range of (1, -1) = -1,

The range of (2, 3) = 3,

The range of (3, 5) = 5,

The range of (4, 8) = 8.

Combining these values we get the range as {-1, 3, 5, 8} which is option D.

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Answer:

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Step-by-step explanation:

let x be the number written to the thousandth place then,

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or, x = 417.91/1000

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Answer:

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3 years ago

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alina1380 [7]

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3 years ago

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Find the absolute extreme of the function on the closed interval. y=9e^xsinx , [0,pi]

Molodets [167]

For convenience sake, I will let $y=f(x)=9e^{x}\sin x$

First, we evaluate the function at the endpoints of the interval.

$f(0)=9e^{0}\sin(0)=0\\\\f(\pi)=9e^{\pi}\sin(\pi)=0$

Then, we need to find the critical points.

We can start by taking the derivative using the power rule.

$f'(x)=9e^{x} \frac{d}{dx} \sin x+9\sin x \frac{d}{dx} e^{x}=9e^{x}(\cos x+\sin x)$

Setting this equal to 0,

$9e^x (\cos x+\sin x)=0$

Since $9e^x > 0$ , we can divide both sides by $9e^x$ .

$\cos x+\sin x=0\\\\\sin x=-\cos x\\\\\tan x=-1\\\\x=\frac{3\pi}{4}$

$f\left(\frac{3\pi}{4} \right)=9e^{3\pi/4}\sin \left(\frac{3\pi}{4} \right)=\frac{9\sqrt{2}e^{3\pi/4}}{2}$

So, the absolute minimum is $\boxed{\left(\frac{3\pi}{4}, \frac{9\sqrt{2}e^{3\pi/4}}{\sqrt{2}} \right)}$ and the absolute minima are $\boxed{(0,0), (\pi, 0)}$

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1 year ago

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svp [43]

Answer:

Step-by-step explanation:

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