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Vedmedyk [2.9K]
3 years ago
13

If n ll m, find the value of x and the value of z.

Mathematics
1 answer:
aliina [53]3 years ago
7 0
Answer: A

Explanation: the two triangles are proportional, so find the “proportion” between the two sides
12/4=3
3*3=9
3z=6
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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
I’m so confused someone please help me
asambeis [7]

Answer:

Figure 5

Step-by-step explanation:

Figure 5, if you draw a line down the middle it is the same on both sides

6 0
2 years ago
Select all expressions that are equivalent to 64
pochemuha

Answer:

Step-by-step explanation:

What do you man what expressions or just expressions in general.

4 0
3 years ago
Which of the folowing is equal to 7 1/3
Romashka-Z-Leto [24]

Answer:

22/3

Step-by-step explanation:

Make the whole number turn into a fraction, (21/3) and add the numerator (1/3), which the answer would be 22/3

7 0
3 years ago
A hummingbird eggs’ mass is approximately 0.374 milligram. What is the mass of 4 Hummingbird eggs?
antoniya [11.8K]

it would be 1.496 milligrams. i hope this helps and i used a calculator

5 0
3 years ago
Read 2 more answers
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