In oM name a chord that lies on a secant

A cube has a volume of 64cm cubed. What is the area of one face of the cube?

goldfiish [28.3K]

Answer:

96 cm squared

Step-by-step explanation:

5 0

3 years ago

A linear function contains the following points.

nydimaria [60]

Find the slope first using m=y2-y1/x2-x1
-4-4/0-4 =2

The y intercept is when x=0.
One of the given points is (0,-4)
The y intercept is y=-4

4 0

3 years ago

I put a picture<br><br> —PLEASE HELP

Karo-lina-s [1.5K]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Base of a cube is a square, and in the given figure it's stated that the measure of each side is " x " ft

Therefore it's Area is equal to : -

$side \times side$

$x \times x$

${x}^{2}$

Now, since it's Area is given as 1/4 ft², let's equate the Areas to find the measure of each side ~

that is ~

${x}^{2} = \dfrac{1}{4}$

$x = \sqrt{ \dfrac{1}{2} \times \dfrac{1}{2} }$

$x = \dfrac{1}{2} \: \: feet$

It's time to find the volume of cube that is ~

$(side) {}^{3}$

So, let's plug the value of side length (1/2 ft)

Volume is equal to ~

${ \bigg(\dfrac{1}{2} \bigg) }^{3}$

$\dfrac{1}{8} \: \: ft {}^{3}$

Hence, the required 1/8 or 0.125 ft³

I hope it helped ~

$\mathrm{✌TeeNForeveR✌}$

3 0

3 years ago

Read 2 more answers

[20 POINTS] A cylinder has a height of 11 centimeters and its circle bases have a radius of 9 centimeters. Find the surface area

galben [10]

The third bubble is correct

4 0

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

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

3 years ago