Help asap!!

The top of a house is 32 feet above ground. The basement floor is 12 feet below

lys-0071 [83]

12 feet below would be -12. from -12 to 0 would be 12 feet.

from 0 to 32 above would be 32 feet.

Total distance = 12 + 32 = 44 feet.

7 0

3 years ago

Three rays have a common vertex on a line. Write and solve equations to find the measures of m and n

lukranit [14]

Answer:

n = 90 - 62

n = 28º

m = 180 - (62 + n + 27)

m = 180 - (62 + 28 + 27)

m = 180 - 117

m = 63º

5 0

3 years ago

Will paid $2.73 for 3.5 pounds of bananas. What was the cost per pound of the bananas?

Alex73 [517]

Answer:

0.78¢

Step-by-step explanation:

if 3.5 lbs of bananas is $2.73, then we need to divide each side by 3.5 to get 1 lb of bananas

1 lb = 0.78¢

To check our work, we can do 0.78x3.5 = 2.73

Hope this helps! Please mark Brainliest if this helped :)

5 0

2 years ago

Select all that apply.

omeli [17]

10:4 15:6 20:6 if those aren’t on there I ain’t ur guy

8 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