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3 years ago

13

Which relation is a function?

1 answer:

igor_vitrenko [27]3 years ago

3 0

Q2c i think that the right one

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Layton has two hamsters.Each hamster drinks 3 centiliters.He fills a water bottle that holds 16 centiliters of water.How many ce

eduard

Answer:

10 centiliters

Step-by-step explanation:

Given data:

2 hamsters are available

1 hamster drinks 3 centiliters

16 centiliters are put

1 hamster drink 3 centiliters therefor 2 hamsters drinks (3 * 2) centiliters = 6 centiliters

<em>Assuming 1 hamster drinks 3 centiliters a day</em>

Remaining centiliters of water will be (16 - 6) = 10 centiliters.

4 0

3 years ago

Let f(x)=5x+2 find f(-6)

LuckyWell [14K]

Answer:

-28

Step-by-step explanation:

Plugging in -6 as x we get 5(-6) + 2, which simplifies to -30 + 2 or -28.

8 0

3 years ago

A 2-digit number is 12 times its tens digit. The sum of the digits is 12. What is the tens digit?

Goryan [66]

C) or 2 is the solution to your question c:

8 0

4 years ago

Read 2 more answers

Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you

Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

With $\pi$ , we should expect $\cos x$ . If $\tan x=\dfrac8{15}$ , then

$\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}$

Also,

$\pi$

so we should expect $\tan\dfrac x2$ and

$\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4$

You seem to be taking

$\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}$

but this is not an identity.

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8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

$\dfrac{\sqrt6-\sqrt2}4$

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9. Use the identity from (7). $\dfrac{7\pi}{12}$ lies in the second quadrant, so

$\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3$

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12.

$\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}$

$=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}$

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

$=\dfrac{\csc x}{\sec x}$

###

13.

$\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

$=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}$

$=\dfrac1{\cos x}$

$=\sec x$

###

14.

$\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)$

$=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}$

$=2$

###

15.

$\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$

$=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$

$=\cos2\theta$

3 0

3 years ago

A rectangular prism has a length of 7 1/2 feet a width of 3 3/4 feet and a height of 2 feet what is the volume of the prism

RideAnS [48]

7 1/2 x 3 3/4 x 2, to get the answer. The answer is

56 1/4 feet squared

3 0

3 years ago