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Butoxors [25]
3 years ago
8

Your friend has $100 when he goes to the fair. He spends $10 to enter the fair and $29 on food. Rude lady at the fair cost $2 pe

r rude. Which function can be used to determine how much money he had left over after x rides?
Mathematics
2 answers:
Aleks04 [339]3 years ago
4 0

Answer:

y=-2x-39

Step-by-step explanation:

storchak [24]3 years ago
4 0

Answer:

100=39+2x

Step-by-step explanation:

hoped that hedlp:P

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Juan bought a piece of lumberthat is 20 feet long.
Semmy [17]

Answer:mutiply not sure

Step-by-step explanation:

7 0
3 years ago
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
Each night, Abby recorded how many pages of a book she read before going to bed. Her data is as follows: 7, 9, 10, 6, 8, 12, 5,
aksik [14]

Answer:

mean = 12.9

median = 9.5

Step-by-step explanation:

mean = sum of all observations / total no. of observations

= 7+9+10+6+8+12+5+11+48+13 / 10

=129 / 10

= 12.9

median = middlemost no. after arranging in ascending order

= 5,6,7,8,9,10,11,13,48

=9 and 10

=9+10 / 2

= 19 / 2

= 9.5

8 0
4 years ago
If l x w = v, what is the value of v?
stepladder [879]
The value of v would be the values of the length times times the width.
3 0
3 years ago
how much cardboard is needed to create a casing that has a base of 12cm, a height of 8cm, and a length of 20cm?
natali 33 [55]
8cm :) of the height :)
7 0
3 years ago
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