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

Solve the following equation for the

Akimi4 [234]

Divide -3
-3/-3a = 15/-3
a = -5

5 0

3 years ago

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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

5 0

3 years ago

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What is the sum of 0.09 and 0.22

iren [92.7K]

I think the answer is 0.31!

7 0

3 years ago

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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