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12345 [234]
3 years ago
6

Help ASAP, please!!!!!!!!!!

Mathematics
1 answer:
Neporo4naja [7]3 years ago
7 0
Can't see it zoom it in a little
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Solve the following equation for the
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Divide -3
-3/-3a = 15/-3
a = -5
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The product of Jack's age and Florence's age is 266. Jack is 14 years<br> old. How old is Florence?
inysia [295]

Answer:

19

Step-by-step explanation:

you divide 266 by 14

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What is the sum of 0.09 and 0.22
iren [92.7K]
I think the answer is 0.31! 
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Please hep me out !!!!!
Volgvan

Answer:

A

Step-by-step explanation:

Assuming you require the distance between the 2 points.

To calculate the distance d use the distance formula

d = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2    }

with (x₁, y₁ ) = (- 3, - 8) and (x₂, y₂ ) = (- 2, - 3)

d = \sqrt{(-2+3)^2+(-3+8)^2}

   = \sqrt{1^2+5^2}

   = \sqrt{1+25}

   = \sqrt{26} → A

7 0
2 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
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