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DiKsa [7]
3 years ago
12

Bought computer for $500. Marked down $50. Plus a 25% discount off original price. What is the original price?

Mathematics
2 answers:
andriy [413]3 years ago
8 0
So 500+50=550

550=25-100=75%

550=75%
multiply both sdies yb 4
2200=300%
divide by 3
733.333=100%

the origonal price=$733.33
vodomira [7]3 years ago
3 0
500 + 50 = 550
500 is 25% of what number?
2000 + 550 = 2550 dollars

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Solve the system by the substitution method.
Ghella [55]

Answer:

A) The solution set is (6,-8).

Step-by-step explanation:

3x - y = 26

-3x       - 3x        Subtract 3x from both sides

-y = -3x + 26     Divide both sides by -1

y = 3x - 26

Now plug this into 7x + 8y = -22 to solve for x

7x + 8(3x - 26) = -22        Distribute

7x + 24x - 208 = -22       Combine like terms

31x - 208 = -22

     + 208    + 208            Add 208 to both sides

31x = 186                           Divide both sides by 31

x  = 6

Plug this into y = 3x - 26 to solve for y

y = 3(6) - 26           Multiply

y = 18 - 26              Subtract

y = -8

If this answer is correct, please make me Brainliest!

8 0
3 years ago
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tino4ka555 [31]

Step-by-step explanation:

it say the answer is 4 but I don't know how

8 0
3 years ago
Integrate this two questions ​
tankabanditka [31]

Simplify the integrands by polynomial division.

\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)

\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)

Now computing the integrals is trivial.

5.

\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}

where we use the power rule,

\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)

and a substitution to integrate the last term,

\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)

8.

\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}

using the same approach as above.

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Answer:

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Step-by-step explanation:

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