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3 years ago

Use the given area of the base and the height to find the volume of the right rectangular prism, in cubic units.

Mathematics

2 answers:

Novosadov [1.4K]3 years ago

8 0

Cant tell because there is no picture

Semenov [28]3 years ago

7 0

You dident give us a picture of it so we cant tell

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What fraction of a hectare is 600 meter squared

nalin [4]

0.06 or 6/100

six hundredths

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3 years ago

A Venn diagram is shown below:

mina [271]

(A ∩ B) means "A intersection B". Your problem statement tells you the intersection contains
{3, 4}

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3 years ago

Read 2 more answers

Does the following table represent a function? Explain.

MAVERICK [17]

Answer:

The table represents a function because each input corresponds to exactly one output.

Step-by-step explanation:

A table can be regarded as representing a function if and only if every input can only be mapped to exactly one output. In other words, it means, only 1 exact output can be assigned to an input. In a table of function, an input cannot have two different outputs. Although, two different inputs can be assigned to the same output.

From the table given, we see that every input has exactly one output. Although, we have different inputs that gives the same output.

Therefore, we can conclude that the table represents a function because each input corresponds to exactly one output.

3 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.

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3 years ago

Which linear inequality is represented by the graph?

Verizon [17]

The answer is the second one, the y-intercept is 3 and its is less than or euqal to so you shade below

5 0

4 years ago