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Charra [1.4K]
3 years ago
12

Find the general solution: y''-2y'-3y=-3te^(-t)

Mathematics
1 answer:
allsm [11]3 years ago
5 0
The answer i got was -3te with -t as an exponent 
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Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
7 0
3 years ago
Five years ago, Tim’s mom was three times Tim’s age today. Now their combined ages are 45. How old is Tim’s mom today?
klio [65]
His mother would be 30.

Hope this helps :)
4 0
3 years ago
Read 2 more answers
Triangle A E C is shown. Line segment B D is drawn near point C to form triangle B D C.
AVprozaik [17]

∠BDC and ∠AED are right angles, is  a piece of additional information is appropriate to prove △ CEA ~ △ CDB

Triangle AEC is shown. Line segment B, D is drawn near point C to form triangle BDC.

<h3> What are Similar triangles?</h3>

Similar triangles, are those triangles which have similar properties,i.e. angles and proportionality of sides.

Image is attached below,
as shown in figure
∡ACE = ∡BCD ( common angle )
∡AED = ∡BDC ( since AE and BD are perpendicular to same line EC and make right angles as E and C)
∡EAC =- ∡DBC ( corresponding angles because AE and BD are parallel lines)

Thus, △CEA ~ △CDB , because of the two perpendiculars AE and BD.

Learn more about similar triangles here:
brainly.com/question/25882965

#SPJ1


3 0
2 years ago
Answer this please for 15 points
spin [16.1K]

Answer:

18/2 : 9

Step-by-step explanation:

4 0
2 years ago
What is the difference between the mean and the median of the data set?
Pepsi [2]
As you probably know there is a difference between mean and median -- the mean acts as a measure of the average, and on the other hand median measures the middle number. The median in this is 15 and the mean is also 15.
4 0
3 years ago
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