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3 years ago

Calculate the volume of the figure.

Mathematics

1 answer:

dimaraw [331]3 years ago

8 0

Answer: The volume is 33 cubic centimeters.

Step-by-step explanation:

First find the volume of the square pyramid on top of the cube. To find the volume of the square pyramid you use the volume a^2*h/3 a is the side length of the base of the square pyramid and h is the height all divide by 3.

So we can say that the side length of the base of the square pyramid is 3 because it has the same side length base as the cube.

V= 3^2 * 2 /3

V= 9 * 2 /3

V= 18/3

v= 6

So the volume of the square pyramid is 6 so now we need to find the volume of the cube and add them together.

Volume of the a cube uses the formula s^3 where s is the side length.

V= 3^3

v= 3*3*3

v= 27

The volume of the cube is 27.

Add 6 and 27 to find the total volume.

6 +27 = 33

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faust18 [17]

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Step-by-step explanation: 21.98 divide by pi is 7 so diameter is 7

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3 years ago

A total of 150 students have taken an Algebra 2 final exam. The scores are normally distributed with a mean of 71% and standard

WARRIOR [948]

Answer:

102 students

Step-by-step explanation:

Note that 65% and 71% are both 1 standard deviation from the mean (71%). According to the empirical rule, 68% of scores lie within 1 std. dev. of the mean.

68% of 150 students would be 0.68(150 students) = 102 students

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3 years ago

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Fittoniya [83]

Answer:

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3 years ago

A radio is on sale for half price. Let y be its original price. Write an expression that tells its sale price.

svetlana [45]

Answer:

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Step-by-step explanation:

4 0

3 years ago

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

3 0

3 years ago