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Dvinal [7]
3 years ago
6

The area of a trapezoid is 39 square millimeters. The height of the trapezoid is 6 millimeters. One of the base legths of the tr

apeziod is 5 millimeters. What is the length of the other base of the trapezoid?

Mathematics
1 answer:
Andrej [43]3 years ago
3 0
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x > 3

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What type of angle is angle G?<br><br> A. obtuse<br> B. straight<br> C. acute<br> D. right
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the correct answer is c

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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
Pemdas? thank you!<br>(-12 ÷ 3) × 2 + 23<br>​
lana66690 [7]

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

( - 12 \div 3) \times 2 + 23 =

( - 4) \times 2 + 23

- 8 + 23 =

23 - 3 - 5 =

20 - 5 = 15

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

3 0
3 years ago
8(C - 9) = 6(20 - 12) - 40​
krek1111 [17]

Answer:

All Real Numbers

Step-by-step explanation:

Any value of <em>c</em> makes the equation true.

Interval Notation

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