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Makovka662 [10]
3 years ago
5

A jewelry store marks up the price of its jewelry up 50%. What will be the selling price of an item that the store purchased for

$95?
Mathematics
1 answer:
snow_tiger [21]3 years ago
3 0

Answer: $142.5

Step-by-step explanation: If the price is being marked up by 50%, you take 50% or half of the original price of $95, which is 47.5, and add it to your original price.

95 + 47.5 = 142.5

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Y= -0.25x + 4.7<br> Y=4.9x - 1.64
Andreas93 [3]

Answer:

The answer would be (1.23, 4.39)

Step-by-step explanation:

Because they are both equal to y, we can set the equations equal to each other and then solve.

4.9x - 1.64 = -0.25x + 4.7

5.15x - 1.64 = 4.7

5.15x = 6.34

x = 1.23

Now that we have the value for x, we can plug into either equation and find y.

y = -0.25x + 4.7

y = -0.25(1.23) + 4.7

y = -.31 + 4.7

y = 4.39

3 0
4 years ago
Rule for raising a power to a new power
KiRa [710]
The practical rule would be, times that number, by the result of the first time you multiplied the first number. But, it would all depend by how much your're raising the number by.

<u>For example:</u> \boxed{4^6}

We do, (4*4*4*4*4*4)= \boxed{\bf{4^6}}= 4096

But, once again, this was an example. This would show and illustrate the rule of "raising a power".<span />
5 0
4 years ago
5/6 = 1/3+d and x/x is a fraction
Alik [6]

\dfrac{1}{3}+d=\dfrac{5}{6}\qquad|-\dfrac{1}{3}\\\\d=\dfrac{5}{6}-\dfrac{1}{3}\\\\d=\dfrac{5}{6}-\dfrac{1\cdot2}{3\cdot2}\\\\d=\dfrac{5}{6}-\dfrac{2}{6}\\\\d=\dfrac{5-2}{6}\\\\d=\dfrac{3}{6}\\\\d=\dfrac{3:3}{6:3}\\\\\boxed{d=\dfrac{1}{2}}

7 0
3 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
3 years ago
Write a function rule for the relationship. the amount c of energy you burn and the time t you spend exercising, if you burn Cal
gregori [183]

Answer:

c= 12t

c is the amount of energy you burn, so it'll be the "solution" to the equation. Multiply 12 times the amount of time you spend exercising to find how much energy you'll burn.

6 0
3 years ago
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