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777dan777 [17]
3 years ago
10

Who the hell is Bucky? ​

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
7 0
I really don’t know....
Sphinxa [80]3 years ago
4 0
I can’t tell if you’re serious or not
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At a Halloween fair, you see a huge Ferris wheel. There are ten seats on the wheel, with two people placed on each seat. Every m
netineya [11]

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answer

Step-by-step explanation:

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3 years ago
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Is Madison have 9876 cookies and milky has 97585 more cookies how many cookies does milky have?
Nina [5.8K]

Answer:

107461

Step-by-step explanation:

9876+97585=107461. If a word problem has the word more it usually means add

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4 years ago
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Plz help and give explaination of how to do it
professor190 [17]

Answer:

25% decreased by 60% is 10

Step-by-step explanation:

New value =

25 - Percentage decrease =

25 - (60% × 25) =

25 - 60% × 25 =

(1 - 60%) × 25 =

(100% - 60%) × 25 =

40% × 25 =

40 ÷ 100 × 25 =

40 × 25 ÷ 100 =

1,000 ÷ 100 = 10

hope this helps :)

7 0
3 years ago
Explain or show work to prove that the following equation is true.6 x 8 = 3 x 2 x 2 x 2 x 2
mr Goodwill [35]

Answer:

see explanation

Step-by-step explanation:

To prove the equation id true, solve both sides and if they are equal then the equation is true.

left side = 6 × 8 = 48

right side = 3 × 2 × 2 × 2 × 2

               = 6 × 2 × 2 × 2

              = 12 × 2 × 2

              = 24 × 2

=             = 48

Since both sides are equal then the equation is true

5 0
2 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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