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andrew-mc [135]
2 years ago
9

F(x)=x-7/x What is the domain of f?

Mathematics
1 answer:
Nesterboy [21]2 years ago
7 0

Answer:

domain: {x ≠ 0}

Step-by-step explanation:

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Answer: To know if a number is repeating it would have this line on top of it, or it would just be repeating. A repeating number is a pattern of numbers that are repeated.

Step-by-step explanation: But 1/4 is not a repeating number, 1/4 is 0.25. This would be a repeating number; 3/9 = 0.333333333...

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Audrey bought one pound of strawberries for $3.20. What is the price, in dollars per ounce of strawberries?
elena-14-01-66 [18.8K]

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There are 16 ounces in 1 pound, so divide $3.20 by 16 to get the price of dollars/ounce

3.20/16 = $0.20/ounce

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2 years ago
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Can someone help me with the question in the image if its right will mark as brainliest
mrs_skeptik [129]

Answer: 14.4

Step-by-step explanation:

x:8.3 = 11.6:6.7

x = 11.6:6.7*8.3 = 14.4

6 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
Simplify this expression. 67 ÷ 65
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Answer:

it should be 67/65 or if in mixed number form :  1  2/65

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