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Art [367]
3 years ago
8

10p^8 divided by 5p^4

Mathematics
1 answer:
REY [17]3 years ago
6 0

Answer:

\sf \cfrac{10p^8}{5p^4}

Factor: \sf 10= 5\times 2

\sf \cfrac{5\times \:2p^8}{5p^4}

Cancel common factor:- 5

\sf \cfrac{2p^8}{p^4}

Use the Quotient Rule:

\sf 8-4=4

\boxed{\sf 2p^4}

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Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).
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Answer:

A) Particular solution:

2x+\frac{1}{2}e^{-x}-\frac{8}{5}

B) Homogeneous solution:

y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))

C) The most general solution is

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

Step-by-step explanation:

Given non homogeneous ODE is

y''+4y'+5y=10x+e^{-x}---(1)

To find homogeneous solution:

D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)

To find particular solution:

y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\

Substituting y_{p},y'_{p},y''_{p} in (1)

y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\

Equating the coefficients

5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\

The general solution is

y=y_{h}+y_{p}

from (2) ad (3)

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

6 0
3 years ago
What is the measure of each side of a square with area equals 225square meter?
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I believe the answer is B
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ohaa [14]

Please find the attachment for a better understanding of the explanation to this question.

As we can see from the diagram attached, the X axis represented by XX' and the Y axis represented by YY' intersect each other at the origin represented by O. Further, we notice that this intersection is at the angle 90 degrees or in other words the intersection is perpendicular. Thus, we have seen that the the x and y axes in the xy plane intersect perpendicularly. And hence, the answer is TRUE.

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