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agasfer [191]
3 years ago
6

2.

Mathematics
2 answers:
Blizzard [7]3 years ago
7 0

Answer:

P' = 1/3 P  

Step-by-step explanation:

Reflection in the y-axis just changes the sign of all the x-values.  Since P" ≅ P', and P' is smaller than P, P" cannot be ≅ to P . It is, however, similar to P.

Good luck!!

Genrish500 [490]3 years ago
3 0

Answer:

P' = 1/3 P

reflection in the y-axis just changes the sign of all the x-values.

Since P" ≅ P', and P' is smaller than P, P" cannot be ≅ to P

It is, however, similar to P.

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Answer:

Step-by-step explanation:

I think the answer should be no because only b is divisible by n not a

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The value of 3√343 is (a) 8 (b)7 (c)6 (d)3​
fredd [130]
The answer will be 3.
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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)
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anygoal [31]

Answer:

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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Though π is a constant

But this question want us to find π

Make π the subject of the formula

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And diameter =12

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Therefore, radius =12/2=6cm

Then

π=P/2r

π=37.7/2×6

π=37.7/12.

Then, the last option is the correct option

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