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!