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3 years ago

11

If −20 is the 15th term in an arithmetic sequence with a common difference of −7, what is the first term in the sequence?

1 answer:

Sphinxa [80]3 years ago

8 0

An=A1+(n-1)d
-20=A1+(15-1)x-7
-20=A1+14 x -7
-20=A1-98
+98 +98
-98+98=0
-20+98=78
A1=78

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Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

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consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

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$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

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$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

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1 year ago

Use Lagrange multipliers to find the maximum and minimum values of

marusya05 [52]

The Lagrangian is

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$L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y$

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From the first two equations we get

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Then

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At these critical points, we have

$f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125$ (maximum)

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3 years ago

A bacteria culture starts with 12,000 bacteria and the number doubles every 50 minutes.

Vlada [557]

Answer:

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b) Approx. 27,569 bacteria

c) About 103 minutes

Step-by-step explanation:

a)

This will follow exponential modelling with form of equation shown below:

$y=Ab^{\frac{t}{n}}$

Where

A is the initial amount (here, 12000)

b is the growth factor (double, so growth factor is "2")

n is the number of minutes in which it doubles, so n = 50

Substituting, we get our formula:

$y=Ab^{\frac{t}{n}}\\y=12000(2)^{\frac{t}{50}}$

b)

To get number of bacteria after 1 hour, we have to plug in the time into "t" of the formula we wrote earlier.

Remember, t is in minutes, so

1 hour = 60 minutes

t = 60

Substituting, we get:

$y=12000(2)^{\frac{t}{50}}\\y=12000(2)^{\frac{60}{50}}\\y=12000(2)^{\frac{6}{5}}\\y=27,568.76$

The number of bacteria after 1 hour would approximate be <u>27,569 bacteria</u>

<u></u>

c)

To get TIME to go to 50,000 bacteria, we will substitute 50,000 into "y" of the equation and solve the equation using natural logarithms to get t. Shown below:

$y=12000(2)^{\frac{t}{50}}\\50,000=12,000(2)^{\frac{t}{50}}\\4.17=2^{\frac{t}{50}}\\Ln(4.17)=Ln(2^{\frac{t}{50}})\\Ln(4.17)=\frac{t}{50}*Ln(2)\\\frac{t}{50}=\frac{Ln(4.17)}{Ln(2)}\\\frac{t}{50}=2.06\\t=103$

After about 103 minutes, there will be 50,000 bacteria

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3 years ago

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nika2105 [10]

Answer:

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Step-by-step explanation:

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