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4 years ago

PLEASEEEEEE HELPPPP

Mathematics

2 answers:

myrzilka [38]4 years ago

7 0

Answer:

C) 14 years

Step-by-step explanation:

Ne4ueva [31]4 years ago

4 0

This is just like an exponential growth equation...

Final=initial*rate^periods

25000=12500(1.0525)^t

2=1.0525^t

ln2=t ln1.0525

t=ln2/ln1.0525

t=13.55

t=14 yrs

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Step-by-step explanation:

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3 years ago

Find the solution of the differential equation that satisfies the given initial condition. y' tan x = 3a + y, y(π/3) = 3a, 0 &lt

Paladinen [302]

Answer:

$y(x)=4a\sqrt{3}* sin(x)-3a$

Step-by-step explanation:

We have a separable equation, first let's rewrite the equation as:

$\frac{dy(x)}{dx} =\frac{3a+y}{tan(x)}$

But:

$\frac{1}{tan(x)} =cot(x)$

So:

$\frac{dy(x)}{dx} =cot(x)*(3a+y)$

Multiplying both sides by dx and dividing both sides by 3a+y:

$\frac{dy}{3a+y} =cot(x)dx$

Integrating both sides:

$\int\ \frac{dy}{3a+y} =\int\cot(x) \, dx$

Evaluating the integrals:

$log(3a+y)=log(sin(x))+C_1$

Where C1 is an arbitrary constant.

Solving for y:

$y(x)=-3a+e^{C_1} sin(x)$

$e^{C_1} =constant$

So:

$y(x)=C_1*sin(x)-3a$

Finally, let's evaluate the initial condition in order to find C1:

$y(\frac{\pi}{3} )=3a=C_1*sin(\frac{\pi}{3})-3a\\ 3a=C_1*\frac{\sqrt{3} }{2} -3a$

Solving for C1:

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Therefore:

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3 0

4 years ago

Please show all steps... finals coming up!

Rama09 [41]

1) find x
y=1/(x+3) -4
y+4=1/(x+3) -4+4
y+4=1/(x+3)

x+3 =1/(y+4)
x= 1/(y+4) - 3

2) change x on y , and y on x

x= 1/(y+4) - 3
y= 1/(x+4) - 3 This is an answer.

4 0

3 years ago