NotThatSmart

Let's plug everything in and go from there. 

Plugging in 4, 2, and -5,

We have \(\sqrt[3]{4(4^4)(2^5)}+\frac{4-(-5)}{(2-5)^2}\)

Simplfying, we get

\(\sqrt[3]{256*32*4}+\frac{9}{(-3)^2}\\ = \sqrt[3]{32768}+1\\ =32+1\\ =33\)

So our final answer is 33. 

Thanks! :)

Let's note something really important first. 

Note that XZW forms a triangle!

Thus, we have

\(\angle ZXW = 90\\ \angle XWZ = 60 \)

Thus, \(\angle XZW = 180 - 90 - 60 = 30\)

So our answer is 30. 

Thanks! :)

Let's note that median of the trapezoid is the average of the two bases. 

12 has to be smaller base, since it's impossible to have negative sidelengths. 

Thus, we can write the equation

\(\frac{12+(12+16)}{2}=40/2 = 20\)

So our answer is 20. 

Thanks! :)

In the first 1000 numbers, there are \(31\) perfect squares, \(10\) perfect squares, and 3 of both. Thus there are

\(1000-(31+10) + 3 = 1000 - 38 = 962 \)

So we have 38 numbers to go! 

We have \(32^2=1024\), so before that we tack on another 23 numbers. We have 15 more numbers. 

The next number we worry about is \(33^2=1089\)

After 1024, the next 15th number is 1039. 

So 1039 is our answer. 

I might be wrong though...

Thanks! :)

Thank you! :)

~NTS

We can set variables to complete this problem quite efiiciently. 

First, let's set the center of the large circle as M and the center of the smaller circle as N. 

Now, let's extend AB and and the line connecting the centers until they meet each other at a pont, O. 

Let the distance of the left edge of the smaller circle and O be x. 

Now, let's note that triangle BMO and ANO are similar. This is quite important. 

From this, we can write

\(BM / MO = AN / NO \\ 4 / (4 + 2 + x) = 1 / ( 1 + x) \\ 4 / ( 6 + x) = 1 /(1 +x) \\ 4(1 + x) = 1 (6 + x) \\ 4 + 4x = 6 + x \\ 3x = 2 \\ x = 2/3\)

Now that we have the value for x, we can easily solve for AB. We get that

\(OA = \sqrt{ NO^2 - NA^2 } = \sqrt{ (1 + 2/3)^2 - 1^2} = \sqrt{ 25/9 -1} = \sqrt{16/9} = 4/3 \\ OB = \sqrt{ MO^2 - MB^2} = \sqrt{ (6 + 2/3)^2 - 4^2 } = \sqrt{ 400/9 -16} = \sqrt{256/9} = 16/3 \\ AB = OB = OA = 16/3 - 4/3 = 12/3 = 4 \)

So our answer is 4

Feel free to let me know if I messed up!

Thanks! :)

We can use an arithmetic sequence to solve this problem. 

The first number in this list is \(1043\), scine 1040 is divisble by 65. 

We just add 65 from there, until we get to 1953, since 1950 is divisble by 65. 

Thus, we have the sequence

\(1043, 1108, 1173...1953\)

Subtracting every term 3 and dividing by 65. we get

\(16,17,18,...30\)

Next, we subtract 15 from every term. 

We get

\(1,2,3...15\)

So our answer is 15...just like what Shmewy said!

Thanks! :)

A terrific number must be a perfect square since it has an odd number of divisors. 

It also has 1 set of divisors other than 1 and itself. 

\(1,4,9\) all don't work since they only have 3 divisors. 

However, \(16\) has exactly 5 factors. We have \(1,2,4,8,16\)

Thus, our answer is 16. 

Thanks! :)

If a number is divisble by 6, then it is also divisble by \(1,2,3,6\)

Now, let's take a look at factors of 6. 

We know that \(12\) has 6 factors, as we have \(1,2,3,4,6,12\)

18 also 6 factors, as we have \(1,2,3,6,9,18\)

However, 24 has exactly 8 factors. We have

\(1,2,3,4,6,8,12,24\)

The prime factors are \(2,3\), so our answer is just 2. 

So there are 2 prime factors. 

Thanks! :)