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3 years ago

11

What is the answer?.....

1 answer:

Butoxors [25]3 years ago

8 0

Answer:

Step-by-step explanation:

f(2) = 2^3 - 4(2)^2 = 8 - 4(4) = 8 - 16 = -8

g(-8)= 4(-8) + 5 = -32 + 5 = -27

C is the answer

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mezya [45]

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3 years ago

What is the sequence rule for : 1.8, 2.58, 3.90, 4.95, 6

slamgirl [31]

I believe there is a typo in the problem. It should be:<span>1.8, 2.85, 3.90, 4.95, 6
If that was the case, the answer will be easy.

In this sequence, the rule is start at 1.80 then keep add 1.05. You can try to put them all into the equation

</span>1.80 + 1.05= 2.85
2.85+ 1.05= 3.90
3.90+ 1.05= 4.95
4.95+ 1.05= 6.00

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3 years ago

45 mi<br> 28 mi<br> What is the length of the hypotenuse?

solniwko [45]

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Step-by-step explanation:

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3 years ago

Identify the line of reflection,<br> O X-axis<br> OY- axis

tigry1 [53]

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4 0

3 years ago

Find all points on the curve x=4cos(t),y=4sin(t) that have the slope of 12.

Marianna [84]

Answer:

$\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$ .

Step-by-step explanation:

We need to find all the points on the curve x=4cos(t),y=4sin(t) that have the slope of 1/2.

$x=4cos (t)$

$\dfrac{dx}{dt}=-4sin (t)$

$y=4sin (t)$

$\dfrac{dy}{dt}=4cos (t)$

Now,

$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$

$\dfrac{dy}{dx}=4cos (t)\times \dfrac{1}{-4sin (t)}$

$\dfrac{dy}{dx}=-\cot t$

So, slope of the curve is $-\cot t$ .

$-\cot t=\dfrac{1}{2}$

$-\tan t=2$

$\tan t=-2$ ...(1)

Using $\sec^2t=1+\tan^2t$ , we get

$\sec^2t=1+(-2)^2$

$\sec^2t=1+4$

$\sec t=\pm \sqrt{5}$

$\cos t=\pm \dfrac{1}{\sqrt{5}}$

Now,

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

$\sin t=\pm \sqrt{1-\dfrac{1}{5}}$

$\sin t=\pm \sqrt{\dfrac{4}{5}}$

$\sin t=\pm \dfrac{2}{\sqrt{5}}$

It equation (1), tan(t) is negative. So, sin and cos have different signs.

If $\sin t= \dfrac{2}{\sqrt{5}}$ , then $\cos t=- \dfrac{1}{\sqrt{5}}$ .

$x=4cos (t)=-\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=\dfrac{8}{\sqrt{5}}$

If $\sin t=- \dfrac{2}{\sqrt{5}}$ , then $\cos t= \dfrac{1}{\sqrt{5}}$ .

$x=4cos (t)=\dfrac{4}{\sqrt{5}}$

$y=4sin (t)=-\dfrac{8}{\sqrt{5}}$

Therefore, the two points are $\left (-\dfrac{4}{\sqrt{5}},\dfrac{8}{\sqrt{5}}\right )\text{ and }\left (\dfrac{4}{\sqrt{5}},-\dfrac{8}{\sqrt{5}}\right )$ .

5 0

3 years ago