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RSB [31]
2 years ago
5

Assignme 3.5.PS-16 : Question Help A shoe store uses a 50% markup for all of the shoes it sells. What would be the selling price

of a pair of shoes that has a wholesale cost of $63?​
Mathematics
1 answer:
Ainat [17]2 years ago
6 0

Answer:

$94.50

Step-by-step explanation:

A shoe store uses a 50% markup for all of the shoes it sells. What would be the selling price of a pair of shoes that has a wholesale cost of $63?​

(63)(1+0.50)=63*1.50=63+31.50=94.50

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2 years ago
11×11=4 22×22=16 33×33=?
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7 0
3 years ago
6x-y=-23 8x+3y=-9 Substition method (Show it step by step please!)
Tanya [424]

Answer:

solution is (- 3, 5 )

Step-by-step explanation:

given the 2 equations

6x - y = - 23 → (1)

8x + 3y = - 9 → (2)

rearrange (1) expressing y in terms of x

y = 6x + 23 → (3)

Substitute y = 6x + 23 into (2)

8x + 3(6x + 23 ) = - 9

8x + 18x + 69 = - 9

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x = - 3

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5 0
3 years ago
Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2
igor_vitrenko [27]

Answer:  The required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Thus, the required solution of the given IVP is

y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

4 0
3 years ago
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