Suzie's Plumbing uses a linear model to determine the total cost, in dollars, of a service call.

vfiekz [6]

Answer:

Could you please expand on your question? Once you do I will edit the answer I put (this) to telling the answer. (The upcoming and this sentence is pre-written and copy and pasted in every brainly question or comment I write or answer.) If this answer helped you please consider giving it brainliest.

- •Trix•

( Yes I know my - •Trix• thingy is not my user but its what I would change it to if I could but I can't, please address me as Trix while commenting or talking to me here.)

8 0

3 years ago

Help please help me pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

horsena [70]

Answer:

x=2/5

Step-by-step explanation:

4 0

2 years ago

Mr. and Mrs. Suralbo earned a net taxable income of P568,986. Find the income tax due.

Contact [7]

Answer:

$198,000

Step-by-step explanation:

Since Mr. and Mrs. Suralbo are married and filing jointly, they would fall into the tax slab of 35% as their taxable income ranges between $414,701 to $622,050.

Taxable income = $568,986

Tax rate = 35%

Income tax due = $568,000 * 35/100

= $198,000

Thus, the income tax due for Mr. and Mrs. Suralbo would be $198,000.

8 0

3 years ago

Seirra found that 15 meters is equivalent to 49.2 feet which of the following is an equivalent ratio

loris [4]

A.........................................

8 0

3 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