hectictar

The diameter of the largest ball that can fit in a cube matches the length of the cube's edge.

volume of a sphere = \(\frac43 * \pi * \text{radius}^3\)

a.

diameter = 1 , radius = 1/2

volume = \(\frac43*\pi*(\frac12)^3=\frac{\pi}6 \approx 0.524 \text{ cubic feet}\)

b.

diameter = 2 , radius = 1

volume = \(\frac43 * \pi * 1^3 = \frac{4\pi}{3} \approx 4.189 \text{ cubic feet}\)

c.

diameter = 3 , radius = 3/2

Do you see what to do now?

If you need more help please just ask :)

AH, I know you would have seen that before long..

What's really ingenious is how you figured out that MrSilvers was trying to take the sin of 15º ...That took some real insight ! 

Hey...that's a really good way of explaining it ! 

\(\cos^2(\frac{\pi}{8})+\cos^2(\frac{3\pi}{8})+\cos^2(\frac{5\pi}{8})+\cos^2(\frac{7\pi}{8}) \\~\\ \cos^2(\frac12\cdot\frac{\pi}{4})+\cos^2(\frac12\cdot\frac{3\pi}{4})+\cos^2(\frac12\cdot\frac{5\pi}{4})+\cos^2(\frac12\cdot\frac{7\pi}{4})\)

Apply the half-angle formula: \( \cos^2(\frac12\cdot a)=\frac12(1+\cos a) \)

\(\frac12(1+\cos\frac{\pi}{4})+\frac12(1+\cos \frac{3\pi}{4})+\frac12(1+\cos\frac{5\pi}{4})+\frac12(1+\cos\frac{7\pi}{4})\)

Now we can evaluate the cosines.

\(\frac12(1+\frac{\sqrt2}{2})+\frac12(1-\frac{\sqrt2}{2})+\frac12(1-\frac{\sqrt2}{2})+\frac12(1+\frac{\sqrt2}{2})\)

Simplify.

\((1+\frac{\sqrt2}{2})+(1-\frac{\sqrt2}{2}) \\~\\ 1+\frac{\sqrt2}{2}+1-\frac{\sqrt2}{2} \\~\\ 2\)

If you want more steps shown please just say so 

width = w

height = w + 28

from the Pythagorean theorem:

w² + (w + 28)² = 52²

w² + (w + 28)(w + 28) = 2704

w² + w² + 28w + 28w + 784 = 2704

2w² + 56w - 1920 = 0

from the quadratic formula:

\(w = {-56 \pm \sqrt{56^2-4(2)(-1920)} \over 2(2)} = \frac{-56\pm \sqrt{18496}}{4}=\frac{-56\pm 136}{4}=-14\pm 34\)

Since we are looking for the length of a line, the answer is positive.

w = -14 + 34 = 20

and

height = 20 + 28 = 48 inches

A Single Lady ? 

arc length = radius * central angle in radians

arc length = 24 * 23π      cm

arc length = 24 * 72.22   cm

arc length = 1733.28      cm

measure of arc across from 39º + measure of arc across from xº = 180º

(2 * 39) + (2 * x) = 180

39 + x = 90

x = 90 - 39 = 51º

You got it! 👍

I can try, but I might get it wrong!

Is this the function: P(n) = 36.5 (1.06)ⁿ

We want to solve for n when P(n) = 70

70 = 36.5 (1.06)ⁿ

70/36.5 = 1.06ⁿ

ln(70/36.5) = ln(1.06ⁿ)

ln(70/36.5) = n * ln(1.06)

\(\frac{\ln(70/36.5)}{\ln(1.06)}=n\approx11.176\)

So..since the function considers the year 2000 the starting point,

The population should reach 70 thousand in the year 2011  

How many "three-fourths" feet of wire are in \( 10\frac12 \) feet of wire?

\(10\frac12 \div \frac34 \\~\\ \frac{21}{2} \div \frac34 =14\)

There are 14 "three fourths" feet of wire in \( 10\frac12 \) feet of wire.

So, there are 14 pairs of earrings that navya is able to make.