How many cubes with side lengths of 1/3 cm does it take to fill a rectangular prism with dimensions 1cm, 2 2/3cm and 2/3cm?

Lyrx [107]

Answer:

Step-by-step explanation:

Okay, so this is impossible to tell the amount of spaces, but let's say it was 4000 for VIP and 4000 for Popular Stands. Multiply $30 by 4000, and you get $120k Earnings. To check your answer, Divide 120,000 by 4000, and you get $30 for each stand.

Now, let's move on with the VIP stands. A VIP Stand costs $50, so 4,000 x $50 = $200k earnings. To check your answer, divide 200,000 by 4000, and you get $50 for each VIP Stand.

If you need to find each variant, go by 1k seat increments, i.e. 1000 and 7000, 2000 and 6000, etc.

Glad I could help!

-Vibilities

8 0

2 years ago

Volume formula of sphere

OleMash [197]

Answer:

Step-by-step explanation:

volume of sphere=4/3(π r^3)

8 0

3 years ago

Read 2 more answers

Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so tha

Pavel [41]

Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

(Expand to obtain an equivalent expression for the sphere: $x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ )

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between $\left(x_1, \, y_1, \, z_1\right)$ and $\left(x_2, \, y_2, \, z_2\right)$ would be:

$\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right)$ .

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .