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3 years ago

Please help me if you can! Thank you (:

Mathematics

1 answer:

AnnyKZ [126]3 years ago

4 0

Answer:

The value of f(2) would be -40.

Step-by-step explanation:

Given function is,

$f(x) = 2x^3 - 3x^2 - 18x - 8$

Substituting x = 2 in the above function,

We get,

$f(2) = 2(2)^3 - 3(2)^2 - 18(2) - 8$

$=2(8) - 3(4) - 36 - 8$

$=16 - 12 - 44$

$\implies f(2)= -40$

Hence, the value of f(2) would be -40.

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Explain in sentences how you can add or subtract two fractions with common denominators?

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Examples:

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$\frac{10}{12} - \frac{2}{12} = \frac{8}{12}$

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3 years ago

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3 years ago

I dont under stand these?

Alexeev081 [22]

The first one means what is the smallest number that goes into 12 which would be 4. The second one mean what does not go into 4 and 6 so that would be 8 because 6 is not a multiple of 8.The third one is either 1 or 2 not sure. And the 4th one is little complicated you might want to look up video on prime factorization that can really help you .

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3 years ago

(I've been trying to figure this out for 3 days and I really need help)

liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

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now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

$\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55$

well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.

pretty much the same thing, we get the volume of the cone and its top, add them up.

$\bf \stackrel{\textit{cone's volume}}{\cfrac{\pi (3)^2(8)}{3}}~~~~+~~~~\stackrel{\stackrel{\textit{half a sphere}}{\textit{top's volume}}}{\cfrac{4\pi 3^3}{3}\div 2}\implies 24\pi +18\pi \implies 42\pi ~~\approx~~131.95~in^$

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3 years ago

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leva [86]

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3 years ago