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3 years ago

It Percent Changepls hurry 30 to 55 (round to the nearest hundredth)

Mathematics

1 answer:

Harlamova29_29 [7]3 years ago

8 0

Answer:

83.33% change

Check work:

30 x 0.8333 = 24.999

30 + 24.999 = 54.999 or about 55

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PLEASE HELP!!!! PLEASE! EASY!! BRAINLIEST RIGHT AWAY IF YOU EXPLAIN!!!

White raven [17]

Answer:

First picture: 60

Second picture:

Step-by-step explanation:

The first step for the first picture is to find the area of the triangle on top. In order to do that, we have to use the equation: (W*L*1/2)

(3*4*1/2)=6

Then, we multiply the above equation by the Height. Which makes the equation become 60.

The first step for the second picture is the same. We first find the area of the triangle on top.

(12*9*1/2)=54

Then we multiply the height by our above answer.

(54*20)=1,080

The other measurements that were there aren't needed. So, next time, completely ignore them!

Hope you have a good day/night! Good luck!

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3 years ago

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PLEASE ANSWER QUICKLY!!The diameter, D, of a sphere is 10.4 mm. Calculate the sphere's volume, V.

Svetlanka [38]

Answer:

About 588.7 mm^3

Step-by-step explanation:

The formula for the volume of a sphere is (4/3)*pi (or in this case 3.14)*r^3, where r is the radius.

We need the radius. We are given the diameter, 10.4.

The radius is half the diameter, so 10.4/2=5.2.

Use r=5.2 and put it into the formula:

(4/3)*3.14*5.2^3≈588.7

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3 years ago

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Which letter below is NOT made only of line segments?<br>D<br>Х<br>М<br>F

MrMuchimi

Answer:

Step-by-step explanation:

The letter D includes a curved line, unlike the letters X, M, and F.

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3 years ago

Of the 25 fish that Karl caught

otez555 [7]

Answer:

1,820 trout

Step-by-step explanation:

Set up a proportion where x is the number of trout in the lake:

$\frac{7}{25}$ = $\frac{x}{6,500}$

Cross multiply and solve for x:

25x = 45,500

x = 1820

So, you could expect 1,820 trout to be in the lake.

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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