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2 years ago

Which expression correctly represents 10 less than a number cubed?

Mathematics

2 answers:

allsm [11]2 years ago

5 0

Answer:

Step-by-step explanation:

$c^{3} - 10$

Art [367]2 years ago

3 0

(x^3)-10

Sorry I can’t see pic

Brainliest is appreciated

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The circumference of a circle is 47.1 and the diameter of the circle is 15. Which best represents

kvv77 [185]

Answer: $\pi =\frac{47.1}{15}$

Step-by-step explanation:

The complete exercise is: "The circumference of a circle is 47.1 and the diameter of the circle is 15. Which best represents the value of π? "

In order to solve this exercise, it is important to remember that the circle can be calculated with the following formulas:

1. $C=\pi D$

Where "C" is the circumference of the circle and "D" is the diameter of the circle.

2. $C=2\pi r$

Where "C" is the circumference of the circle and "r" is the radius of the circle (Remeber that the diameter is twice the radius).

In this case, the exercise gives you the circumference of the circle and its diameter. These are:

$C=47.1\\\\d=15$

Then, knowing those values, you can substitute them s into the first equation $C=\pi D$ , as following:

$47.1=\pi (15)$

The final step is to solve for $\pi$ :

$\pi =\frac{47.1}{15}$

8 0

2 years ago

Solve for x<br><br> C = 1/4πs^2x

kvasek [131]

Answer:

x=4C/πs^2

Step-by-step explanation:

C=1/4πs^2x

1) Divide both sides by π and s^2:

C/πs^2=1/4x

2) Multiply both sides by the reciprocal of 1/4: (4)

4C/πs^2=x

6 0

2 years ago

A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and

natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: $E=k\frac{q}{r^{2} }$ ; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: $p=\frac{q}{A}$ ; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: $S=4\pi r^{2}$ ; which gives us the surface of a sphere.