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A computer code written in C++ finds that there are:

9,000 7-digit palindromes

Of the above 9,000 palindromes:

2,000 of them are multiples of 4

An equilateral triangle means:
It has 3 equal sides and 3 equal angles of 60 degrees each. Since each angle of the larger square ==90 degrees, then the 2 smaller angles on either side==[90 - 60] / 2 ==15 degrees.

Now, you have an Obtuse isosceles triangle on each side of the larger square with angles of 15, 15 and 150 degrees with two equal sides of 2 units each.

Using the Law of Cosines, we therefore find the 3rd side of Obtuse Isosceles triangle to be:

c==Sqrt[2^2+2^2 - (2*2*2*cos(150)) ]
c==~3.864 units - which is also the side of the larger square.

Hey I don't think 24 would be the correct answer
Also do you think there is another way to do it? I've never seen this before.

This is calculated using "Stirling Numbers of the Second Kind"

k=4; n=4; (1/k! * sumfor(i, 0, k,(-1)^i * (k nCr  i) * (k - i)^n) * k!

==24 different ways - without any restrictions

I saw this same question answered ealier today.

Take a look for it.

Like this for Q1:

Now let's see you attempt something!

never mind, I got it

If you go on to a scientific calculator, 99^154 = 2.12725703×10^307 but then 99^155 is Undefined, so the anser is 154.

154

In whose universe is 7 the difference between 50 and 52 ?  

(4+4-1) C (4-1) == 35 ways - with no restrictions

See https://web2.0calc.com/questions/the-fundamental-of-calc-pt1#r1, for the approach.

Split the integral into two integrals (note the upper limit on the first integral and the lower limit on the second integral):

\(\int_{-1}^{2}f(x)dx=\int_{-1}^{1}(3-x^3)dx+\int_{1}^{2}x^2dx \)

I assume you can do these individual integrals.

How much money did Phil have when he started?

Hello Guest!

\(x-100-\frac{x-100}{2}-50-20-\frac{x-100-\frac{x-100}{2}-50-20}{2}-3=0\\ 2x-200-x+100-100-40-x+100+x-100-70-3=0\\ \color{blue} x=313\)

313$ did Phil have when he started.

  !

Omg lol. I wrote that down but i didn't know if that was correct and if i didn't know if i had to do more with that lol. Thanks

Just count em...

4 - 0 - 0 - 0 

3 - 1 - 0 - 0

2 - 2 - 0 - 0

2 - 1 - 1 - 0

1 - 1 - 1 - 1 

So \(\color{brown}\boxed{5}\) ways

answered on the original

Each ball has 4 choices, so there are \(4^4 = \color{brown}\boxed{256}\) ways.