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3 years ago

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A store is having a 15% off sale on all shirts. When Suri checks out, the clerk takes $4.50 off the original price of the shirt.

What was the original cost of the shirt? Use the formula to find the answer.
15% of x = $4.50

1 answer:

Nastasia [14]3 years ago

5 0

Answer: $30

Step-by-step explanation:

Starting with the original formula, you can solve for the cost of the shirt.

First, convert the percentage to a decimal - 15% = .15, and then plug it into the formula:

.15x = $4.50

In order so solve this you need to divide both sides by .15:

X = $30

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Solve these linear equations in the form y=yn+yp with yn=y(0)e^at.

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a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

b) $y(t) = y_{0}e^{-4t} + 2$ . It has a steady state.

Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

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$4C = 8$

$C = 2$

The solution in the form is

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$y(t) = y_{0}e^{4t} + 2$

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The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

So

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

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