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3 years ago

What is true about the function graphed below?

Mathematics

1 answer:

Dafna1 [17]3 years ago

8 0

Answer:

D. The axis of symmetry is the y-axis.

Step-by-step explanation:

The vertex is at coordinates (0, 3). Since this parabola opens vertically, its axis of symmetry is the x=coordinate of the vertex: x = 0. That is the equation of the y-axis.

The axis of symmetry is the y-axis.

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DENIUS [597]

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Step-by-step explanation:

base² + Altitude² = Hypotenuse²

x² + 9² = 13²

x² + 81 = 169

x² = 169 - 81

x² = 88

x = $\sqrt{88}$

$x = \sqrt{2*2*2*11}=2\sqrt{2*11}\\\\x=2\sqrt{22}$

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2 years ago

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20 divided by 2 (5+5)=

AveGali [126]

<span>The correct answer is 1.

Explanation:
The way this is written, everything after the words "divided by" would be calculated first; it is written as though this is a grouping, such as 20/(2*(5+5)). We would evaluate the innermost parentheses first; 5+5=10. Then this gives us 2*10=20 in our parentheses. 20/20 = 1.</span>

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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