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algol13
3 years ago
10

Thomas bought a new backpack that was 30% off the original price. The sale resulted in a $15 discount. What was the original pri

ce of the backpack?
a.) $45
b.)$50
c.)$65
d.)$70
Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
5 0
30% = $15

1% = $15 ÷ 30 = $0.50

100% = $0.50 x 100 = $50

Answer: $50 (Answer B)
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You are selling tomatoes. You have already earned $16 today. How many additional pounds of tomatoes do you need to sell to earn
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2 years ago
Divide 310AED among A B C in the ratio 2:3:5
Shkiper50 [21]

Answer:A=62AED B=93 C=155

Step-by-step explanation:

You add all of the numbers to split between (2,3,5) you get 10

then you divided 310 by 10 which is 31

then you times 2 by 31=62 (a)

3x31=93 (b)

5x31=155 (c)

5 0
2 years ago
Read 2 more answers
Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.
Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If x + \dfrac{1}{x} is an integer

∴ \left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right ) is an integer

From which we have;

x^3 + \dfrac{1}{x^3} is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

x + \dfrac{1}{x}

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}

By simplification of the cube of the given integer expressions, we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )

Therefore, we have;

\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}

By rearranging, we get;

x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )

Given that  x + \dfrac{1}{x} is an integer, from the closure property, the product of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 is an integer and 3\cdot \left (x + \dfrac{1}{x} \right ) is also an integer

Similarly the sum of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right  ) is an integer

\therefore x^3 + \dfrac{1}{x^3} =   \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right  ) is an integer

From which we have;

x^3 + \dfrac{1}{x^3} is an integer.

4 0
3 years ago
Find the total area of the kitchen floor that is to be tiled as shown below.
schepotkina [342]

Answer:

58

Step-by-step explanation:

There is two rectangles in this problem.

For the first one 5×10=50

And the second one is 2×4=8

So the final answer will be 58

Hope this helpedn

7 0
3 years ago
Read 2 more answers
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