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Naya [18.7K]
3 years ago
12

|8 - 7| · 4 + 3(-1) use pemdas

Mathematics
2 answers:
mamaluj [8]3 years ago
5 0
<span>|8 - 7| · 4 + 3(-1)
= 1 </span>· 4 - 3
= 4 - 3
= 1
hope it helps
MatroZZZ [7]3 years ago
4 0

pemdas

P - parentheses

E - exponents

M - multiplication

D - division

A - addition

S - subtraction

|8 - 7| · 4 + 3(-1)

= |8 - 7| x 4 + -3

= 8 - 7 x 4 + -3

= 8 - 28 + -3

= 8 - 25

= -17

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From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

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y_1=e^x\cos x
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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
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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
Brainliest to right person
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Answer:

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So, we now that 4 ÷ 6/9 is equal to 4 x 9/6. 4 x 9/6 = (4x9)/6 = 36/6 = 6.

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Instead of typing it all on here, I included my explanations with the work.

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