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fiasKO [112]
2 years ago
7

PLEASE HELP :((

Mathematics
1 answer:
Mazyrski [523]2 years ago
3 0

Answer

32 is the answer....

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Please help me with this please.
Leno4ka [110]

Answer:

x is greater or equal to 7

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
A 20-ounce soda cost $1.20. How much does the soda cost per ounce?​
Andre45 [30]

Answer:

Your answer is 0.06, hun!

Step-by-step explanation:

1.20 divided by 20 = 0.06

Stay happy-Livia

3 0
2 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
7 0
3 years ago
Plz help marking brainiest answers!!!
Nady [450]
2(13) = 26
-3(6) = -18
26 -  18
=8
so the answer is A
8 0
3 years ago
F(x)=2x+1/x-1 then f^-1(1)​
olasank [31]

Answer:

  • f⁻¹(x) = (x + 1) / (x - 2)
  • f⁻¹(1 ) = - 2

Step-by-step explanation:

<u>Given function:</u>

  • f(x) = (2x + 1) / (x - 1)

<u>Find its inverse, substitute x with y and f(x) with x, solve for y:</u>

  • x = (2y + 1) / (y - 1)
  • x(y - 1) = 2y + 1
  • xy - x = 2y + 1
  • xy - 2y = x + 1
  • y(x - 2) = x + 1
  • y = (x + 1) / (x - 2)

<u>Substitute y with f⁻¹(x):</u>

  • f⁻¹(x) = (x + 1) / (x - 2)

<u>Find  f⁻¹(1 ):</u>

  • f⁻¹(1 ) = ( 1 + 1) / (1 - 2) = 2 / - 1 = - 2
6 0
2 years ago
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