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sweet-ann [11.9K]
3 years ago
12

I need help with math

Mathematics
1 answer:
aksik [14]3 years ago
8 0
Well what is the problem ?
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Describe how 0.05 and 0.005 compared
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0.05 is on the place of hundred and 0.005 is on the place of thousand
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91403 divided by 21. Show work
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The given Divisor = 21 and Dividend = 91403

43522191403847463110105534211

The Quotient is 4352 and the Remainder is 11

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Plas answer this I need it today
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17 : Both

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2 years ago
3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
Which inequality statement describes the two numbers on a number line? "−5 and a number 4 units to the right of −5" A) 4 < −5
Marina CMI [18]

Answer: D

Step-by-step explanation:be cause you have to add 4 to negetive 5 and and that negetive 1 which is greater than -5

8 0
3 years ago
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