1. What is the value of x? Show all of your work.

1 answer:

4 0

Pythagorean Theorem: A^2 + B^2 = C^2
We know A and C, let's find B by doing it backwards 

C^2 - A^2 = B^2 
4^2 - 7 (square root of 7 squared is just 7) = B^2 
16 - 7 = B^2 
9 = B^2

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SOMEONE PLEASE HELP MEEEE

AURORKA [14]

Answer:

$-180^{\circ}$

Step-by-step explanation:

Lines form 180 degrees on either side. The negative sign simply implies direction (clockwise instead of counter-clockwise), which is negligible as 180 is exactly half of 360 so you'd form the same line.

8 0

2 years ago

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What is the assessed valuation of a house with a $67,170 market value and a 49 percent assessment rate?

-BARSIC- [3]

Had to look for the missing options and here is my answer.

Given that the market value of the house is $67,170 and it has a 49 percent assessment rate, the assessed valuation of this house would be $32,913. So how we got this amount is that we multiplied 0.49 by 67,170 and we got 32,913.3. Hope this answers your question.

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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