I would really appreciate it if someone could answer these as i'm really stuc and its due for tomorrow.

1 answer:

7 0

1. Round 50.75 to 50 and 0.18 to 0.20.
50 x 0.2 = 10

2) Round 96 to 100 and 0.499 to 0.5
100 divided by 0.5 = 50

3a) Round 8.2 to 8, 6.7 to 7, and 0.46 to 0.50
8 x 7 divided by 0.50 = 112

3b) Round 23.4 to 20, 13.9 to 10, and 0.18 to 0.20
20 x 10 divided by 0.20 = 2,000

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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

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

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Yesny had $240 to spend on five pairs of jeans. How much did each pair of

krek1111 [17]

Answer:

5x + 10 = 240

Step-by-step explanation:

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3 years ago

Read 2 more answers

I need an explanation how to do this ASAP I will give you 20 POINTS!!

mars1129 [50]

For the first digit you are choosing from 2 digits
for the second digit you are choosing from 2 digits
for the third digit you are choosing from 2 digits
2*2*2=8
111
110
011
101
000
001
100
010
However......
A number doesn't usually start with a 0 or 0s.
Therefore, if you want 3-digit numbers and not just permutations using 0 and 1, then you must eliminate
011, 000, 001, and 010
Seeing that the first digit can't be 0, you choose from 1, then 2, and then 2 digits again; 1*2*2=4 numbers
You choose which answer best suits your problem.

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Galina-37 [17]

The answer is y=-8 I’m pretty sure

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posledela

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3 years ago