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2 years ago

Zack and Noah jogged 2/5 of a mile and walked another 1/4 of a mile. How far did they go in all?

1 answer:

3 0

Find a common denominator. The lowest one would come from you multiplying 2/5 by 4 on each side and 1/4 by 5 on each side. You are left with 8/20 and 5/20, added together would be 13/20.

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Twelve divided by the sum of x and 2 equals the quotient of 4 and the difference of x and 2. Find x.

Marizza181 [45]

Answer:

Step-by-step explanation:

12/(x+2) = 4/(x-2)

Cross-multiply to get:

12x-24 = 4x+8

8x = 32

x = 4

7 0

3 years ago

Completely factor the given expression over the integers.<br><br> (Correct gets brainliest)

elena-14-01-66 [18.8K]

Answer:

Step-by-step explanation:

would it be 50 because you the q to the power of 2 is 2q. Then you would subtract to the other side of the equation so it would be -100=-2q. After you would divided the -2 in -2q on both sides and -100/-2 is 50. So your answer is q=50

7 0

3 years ago

Find f of 1 and f of 4

Dafna1 [17]

You have to check f(1) in each one to see if it's right and for f(4)

4 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