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Nataly [62]
3 years ago
11

Jessa bought her home for $125,000 in 2010 Property values have increased 15% every year since she has owned the home. Which of

the following equations can be used to represent the price of the home x years after 2010?
y = 125,000(1 15)
y = 125 000(125)
y = 125,000(085)
y = 125.000(075)
Mathematics
2 answers:
kramer3 years ago
3 0

Answer:

y= 125,000 (1.15x)

Step-by-step explanation:

timurjin [86]3 years ago
3 0

Answer:

A. y=125,000\cdot (1.15)^x

Step-by-step explanation:

We have been given that Jessa bought her home for $125,000 in 2010 Property values have increased 15% every year since she has owned the home.

We can see that increase in value of house is not constant, so the value of house in increasing exponentially.

We know that an exponential function is in form y=a\cdot b^x, where,

a = Initial value,

b = For growth b is in form (1+r), where r represents growth rate in decimal form.

r=\frac{15}{100}=0.15

y=125,000\cdot (1+0.15)^x

y=125,000\cdot (1.15)^x

Therefore, the equation y=125,000\cdot (1.15)^x represents the price of the home x years after 2010.

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Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular soluti
Vikki [24]

Answer:

General Solution is y=x^{3}+cx^{2} and the particular solution is  y=x^{3}-\frac{1}{2}x^{2}

Step-by-step explanation:

x\frac{\mathrm{dy} }{\mathrm{d} x}=x^{3}+3y\\\\Rearranging \\\\x\frac{\mathrm{dy} }{\mathrm{d} x}-3y=x^{3}\\\\\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{3y}{x}=x^{2}

This is a linear diffrential equation of type

\frac{\mathrm{d} y}{\mathrm{d} x}+p(x)y=q(x)..................(i)

here p(x)=\frac{-2}{x}

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The solution of equation i is given by

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we have e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}

Thus the solution becomes

\tfrac{y}{x^{2}}=\int \frac{1}{x^{2}}\times x^{2}dx\\\\\tfrac{y}{x^{2}}=\int 1dx\\\\\tfrac{y}{x^{2}}=x+cy=x^{3}+cx^{2

This is the general solution now to find the particular solution we put value of x=2 for which y=6

we have 6=8+4c

Thus solving for c we get c = -1/2

Thus particular solution becomes

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5 0
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Answer:

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Step-by-step explanation

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∴ radius  r=\sqrt{(0-0)^{2}+(5-0)^{2} }

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∴ The equation of the circle is

     x^{2} +y^{2} = 25

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Step-by-step explanation:

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