All categories

3 years ago

Hey please please please please please help have a Struggling with this all day

Mathematics

1 answer:

Inessa05 [86]3 years ago

6 0

If the side lengths are all 6 units, then the surface area and volume are the same

surface area = 6*s^2 = 6*6^2 = 216
volume = s^3 = 6^3 = 216

You can find this by solving s^3 = 6s^2 for s to get s = 0 or s = 6. The solution s = 0 is trivial so you can ignore it.

You might be interested in

Which expresión Is true and could you explain it

Verdich [7]

First equation
L.H.S = 4x -0.25y
so it doesn't equal the R.H.S

2nd equation
R.H.S = 15q + 3qr
so it doesn't equal the L.H.S

3rd equation
R.H.S = -30m -60n
so R.H.S doesn't equal L.H.S

4th equation
L.H.S = -16a +4
so L.H.S equal R.H.S

The answer is D

8 0

3 years ago

Rhonda is making an input/output table.When she sees the number 5,she writes 3.0.When she sees 10,she writes 6.0.When she sees 4

zhannawk [14.2K]

7.2 and 7 are the answers

4 0

3 years ago

Two equivalent decimals for 5.3

Eduardwww [97]

Two equivalent decimals for 5.3 are 5.30 and 5.300

6 0

3 years ago

Y = 10.5 x + 5<br> Please help me solve this equation.<br> Solve for x.

Schach [20]

What is the context of the question

8 0

3 years ago

The solutions of the quadratic equation Ax^2+Bx+C=0 are called roots and are given by a famous equation called the Quadratic For

schepotkina [342]

The roots of a quadratic equation depends on the discriminant $\Delta$ .

If $\Delta > 0$ , the quadratic equation has two real distinct roots, and it crosses the x-axis twice.
If $\Delta = 0$ , the quadratic equation has one real root, and it touches the x-axis.
If $\Delta < 0$ , the quadratic equation has two complex roots, and it neither crosses nor touches the x-axis.

---------------------

A quadratic equation has the following format:

$y = ax^2 + bx + c$

It's roots are:

$y = 0$

Thus

$ax^2 + bx + c = 0$

They are given by:

$\Delta = b^{2} - 4ac$

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

The discriminant is $\Delta$ .

If it is positive, $-b + \sqrt{\Delta} \neq -b - \sqrt{\Delta}$ , and thus, the quadratic equation has two real distinct roots, and it crosses the x-axis twice.

If it is zero, $-b + \sqrt{\Delta} = -b - \sqrt{\Delta}$ , and thus, it has one real root, and touching the x-axis.

If it is negative, $\sqrt{\Delta}$ is a complex number, and thus, the roots will be complex and will not touch the x-axis.

A similar problem is given at brainly.com/question/19776811

4 0

2 years ago