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murzikaleks [220]

3 years ago

9

You deposit $20,000 in an account that pays 7.77% annual interest. Find the balance after 2 years when the interest is compounde

d monthly.
The balance after 2 years is $.

1 answer:

zvonat [6]3 years ago

3 0

Answer:

=$23,350.802

Step-by-step explanation:

Compound Interest (A)=P(1+r/n)^nt

Where,

P=principal=$20,000

r=Interest rate=7.77%=7.77/100

=0.0777

n=number of periods=12 month in a year

t=time=2 years

(A)=P(1+r/n)^nt

=$20,000(1+0.0777/12)^12×2

=$20,000(1+0.006475)^24

=$20,000(1.006475)^24

=$20,000(1.1675401)

=$23,350.802

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Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

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3 years ago

Plz helppp I’m not so smart:)

olya-2409 [2.1K]

Step-by-step explanation:

I think you will get x this way:

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3 years ago

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grigory [225]

The graph of f(x) + 1 is the graph in the option C.

<h3>Which is the graph of f(x) + 1?</h3>

For a given function f(x), a vertical translation is written as:

g(x) = f(x) + N

If N > 0, then the translation is upwards.
If N < 0, then the translation is downwards.

Here we have g(x) = f(x) + 1, so we have a translation of 1 unit upwards, the graph of f(x) + 1 is the graph of f(x) but translated one unit upwards.

From that, we conclude that the correct option is C.

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2 years ago

Brandon buys a radio for $43.89 in a state where the sales tax is 7percent

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Answer:

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