tertre

We know that the area of the shaded region plus the area of circle A equals the Area of Circle B. Since the shaded area: area of the circle, 3:1, we have $3x+x=64\pi$$4x=64\pi, x=16\pi$ . Directly, we get  $\pi*r^2=16\pi$, so $r^2=16, r=4.$ Remember that length can't be negative! Thus, the diameter of circle A is $4*2=\boxed{8}$ cm.

Yep, try to isolate the like terms! 

We have $\frac{2000}{24}=83 R 8$ , so   $8$ hours from $2:00 A.M.$ should be $\boxed{10A.M.}.$

2. Rewriting this, we have: $3x+3=4y+3, \frac{3x}{4}=y.$ Now, plugging a value of $x$ that will leave $y$ are a non-negative integer, is $x=4.$ Answer is 3? I might be wrong...

$2^2=\boxed{4}.$

We simplify: $\left|-2\left(-5\right)-38\right|+\frac{2}{3}\left(-9+7\right)^3=\left|2\cdot \:5-38\right|+\frac{2}{3}\left(-9+7\right)^3$, so we get $28+\frac{2}{3}\left(7-9\right)^3$ or $\boxed{\frac{68}{3}}.$

1. Imagine the side lengths of the square are (s, s, s, s) . The area is the same since s*s=s^2 Now, label the figures correctly and you would see that AC is the diagonal of the square, which is $s\sqrt{2}.$ Now, we have for rectangle AC*CE=$s^2$, but we already know AC, so we have $CE=\frac{s^2}{s\sqrt{2}}$ . Rationalizing the denominator, we get $CE=s\sqrt{\frac{5}{2}}$ . Now, dividing AC by CE, we get an answer of $\frac{2}{\sqrt{5}}$ , which can be changed into   $\boxed{\frac{2\sqrt{5}}{5}}.$

Do you have a diagram for problem 2?

Hope this helped,

tertre.

Imagine a right triangle: We have adjacent/hypotenuse, so 20 cos 72 approx 6.2! Like what EP had said!

You can add them up by grouping keenly: 3+6+9+2+5+8+1+4+7+0+3+6+9+2+5+8, pair 3 with the 7, 6 with the 4, 9 with the 1 and so on. Finally, you'll get $\boxed{78}.$

I'm fine, thank you!