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uysha [10]
3 years ago
13

Solve the equation what is X+7=-8

Mathematics
1 answer:
tino4ka555 [31]3 years ago
6 0

Answer

-15 + 7 = -8

Explanation

-15 + 7 = -8

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Please help if you can and if you know how to help me understand these Im all ears.
maw [93]

Aight let's answer this step by step

What you need to know is if two values are equal either one will be similar to the other it's only the format that's different if you calculate each of the fractions you'll see each pair is equal except for the last option

Where if calculated

8/3=2.67 but

5/4.8=1.04

Therefore values aren't equal to each other

6 0
2 years ago
Read 2 more answers
Determine each segment length in right triangle ABC
zepelin [54]

Answer:

BC = 7√2

BD = 7

Hope this helps you.....

6 0
2 years ago
One of my last ones i need
grandymaker [24]

angle EPF = angle DPG

=> 4x + 48deg = 7x

=> 3x = 48 deg

=> x = 16

=> angle EPF = 4(16) +48 = 112deg

7 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
If x^2=40, what is the value of x?
Ainat [17]

if x^2=40 you would need to square root both sides to get x alone therefore x=6.3

5 0
3 years ago
Read 2 more answers
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