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notka56 [123]
3 years ago
7

Is this a no solution answer2(4x+1)-3=Ax+b

Mathematics
1 answer:
butalik [34]3 years ago
8 0

Answer:

8x+2-3=Ax+b=x=8-2-3=7

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Compute the slope of the line below. Enter your answer as a fraction or
Arada [10]

Answer:

  1. yirsurusitr

Step-by-step explanation:

f

  1. fftzethatwtk
  2. dg5teyd5lwey Hrhht c.f.rsxtmrshzt fsfyrscytn

5 0
3 years ago
Which of the following is the solution to | x| -5< -13 ?
sergey [27]
|x| - 5 < -13
|x| < -13 + 5
|x| < -8

No solution.
6 0
3 years ago
Solve for x..........................
Lelechka [254]

Answer:

3

Step-by-step explanation:

If your talking about the x where it says 3x in, then

3x = 9

x = 3

4 0
3 years ago
What is 18 7/8 -14 7/9
strojnjashka [21]

Answer:

295/72

Step-by-step explanation:

6 0
3 years ago
Solution for dy/dx+xsin 2 y=x^3 cos^2y
vichka [17]
Rearrange the ODE as

\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y
\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3

Take u=\tan y, so that \dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}.

Supposing that |y|, we have \tan^{-1}u=y, from which it follows that

\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}
\sec^2y=1+\tan^2y=1+u^2

So we can write the ODE as

\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3

which is linear in u. Multiplying both sides by e^{x^2}, we have

e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}
\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}

Integrate both sides with respect to x:

\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx
e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx

Substitute t=x^2, so that \mathrm dt=2x\,\mathrm dx. Then

\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt

Integrate the right hand side by parts using

f=t\implies\mathrm df=\mathrm dt
\mathrm dg=e^t\,\mathrm dt\implies g=e^t
\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)

You should end up with

e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C
u=\dfrac{x^2-1}2+Ce^{-x^2}
\tan y=\dfrac{x^2-1}2+Ce^{-x^2}

and provided that we restrict |y|, we can write

y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)
5 0
4 years ago
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