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Ludmilka [50]
3 years ago
10

Suppose you are offered just one slice from a round pizza (in other words, a sector of a circle). The slice must have a perimete

r of 32 inches. What diameter pizza will reward you with the largest slice? Specify the diameter and the maximum area of the slice.
Mathematics
1 answer:
Andrew [12]3 years ago
6 0
Dimeter is 28inch and area is equal to 26.7inch^2
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Put these fractions in order from least to greatest:
rewona [7]

Answer:

1/4, 3/8, 2/3

Step-by-step explanation:

3/8 is more than 25% (1/4) but is less than 66% (2/3) making the order 1/4 3/8 2/3.

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Find the equation of the line that passes through (-1,-3) and is perpendicular to
miv72 [106K]

Answer:

y = -2x - 5

Step-by-step explanation:

2y = x - 3

y = 1/2x - 3/2

gradient = 1/2

perpendicular gradient = negative reciprocal = -2

y = -2x + c

(-3) = -2(-1) + c

-3 = 2 + c

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2 years ago
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What is the answer ?<br> x/y+17 <br> If is x=12 <br> and y=1/2
zavuch27 [327]

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jhhdh.duuemmeyyeymruur*

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Cartoon=uuu
LekaFEV [45]

Answer:

three u's and an I

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3×3=9

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8 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
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