Please help me)) I need help with this!

Mathematics

1 answer:

melomori [17]3 years ago

6 0

Answer:

5? im not sure

Step-by-step explanation:

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OverLord2011 [107]

The answer:

Is d have a good day

5 0

4 years ago

angle 0 is in quadrant 1 with sin0 = 2/5. Use the Pythagorean identity, sin^2 0 + cos^2 0 = 1, to calculate the value of cos0 as

Gwar [14]

We know that
sin²x+cos²x=1
so
clear cos x
cos x=(+/-)√[1-sin²x]

in this problem
Angle 0 is in quadrant 1 -----> cos o and sin o are positive
sin o=2/5
cos x=√[1-(2/5)²]----> cos o=√[1-4/25]----> cos o=√[21/25]---> cos o=√21/5

the answer is
cos o=√21/5

7 0

3 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

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