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Delvig [45]
3 years ago
15

DO NOT PUT IN A LINK JUST TELL ME this is serious this is a test(NOO LINK!!!!!!)

Mathematics
1 answer:
Svet_ta [14]3 years ago
7 0

Answer:

the answer is 4 because 16×4 is 64

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Interpreting Solutions:
Darina [25.2K]
We have that
solution is
x <-10 or x>=25

for
x < -10
the solution is the interval (-∞, -10)

and 
for 
x>=25
the solution is the interval [25,∞)

therefore
the solution for
x <-10 or x>=25
is
(-∞, -10)  ∪ [25,∞)

<span>the solution is any value of x less than -10 or greater than or equal to 25</span>
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Answer:

bjrkhfjehrukfe

Step-by-step explanation:

rnfewklnfeklr

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Can someone help me please answer this question
andre [41]
I’m pretty sure the answer would be Y=42°
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4 years ago
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Which equation shows that 6 is a factor of 18?
crimeas [40]
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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
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