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Artist 52 [7]
3 years ago
9

An investor opened an account with an online brokerage firm and deposited $1,000.00 into the account. She then used all the mone

y in the account to buy some shares in a company, costing her $15.00 in fees, and made a 10% profit on the remainder of her money. She then sold the shares, costing her $15.00 in fees, and bought some shares in another company with the remainder of her money, costing her $15.00 in fees, and she made a 10% profit on the remainder of her money. Finally, she sold the shares, costing her $15.00 in fees. Which of these recursive formulas will produce the sequence that represents her original number of dollars, her number of dollars after her first sale, and her number of dollars after her second sale?
Mathematics
1 answer:
IgorC [24]3 years ago
7 0
T1 = 1000, tn = (1.1)(tn – 1 – 15) – 15
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Answer:

The solution is y=-\frac{12}{12x-30x^2+1}.

Step-by-step explanation:

A first order differential equation y'=f(x,y) is called a separable equation if the function f(x,y) can be factored into the product of two functions of x and y:

f(x,y)=p(x)h(y)

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We have the following differential equation

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In the given case p(x)=1-5x and h(y)=y^2.

We divide the equation by h(y) and move dx to the right side:

\frac{1}{y^2}dy\:=(1-5x)dx

Next, integrate both sides:

\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C

Now, we solve for y

-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}

We use the initial condition y(0)=-12 to find the value of C.

-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}

Therefore,

y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}

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