CubeyThePenguin

Let the numbers be x and y.

one number is four less than a second number ==>    x = y - 4         . (1)

4x the first is 14 more than 5x the second ==>     4x = 5y + 14          . (2)

We will use substitution. Plug equation (1) into (2):

4(y - 4) = 5y + 14

4y - 16 = 5y + 14

-y = 30

The numbers are 30 and 26.

\((\frac{1}{3})^3 = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \boxed{\frac{1}{27}}\)

Double repost. I answered 2 and 3 here:

A pentagon can be rotated 5 ways and still have it look the same. I hope this helps. I myself still am not sure of the solution yet.

This sum is 1 + 2 + 4 + 8 + 16 = 31.

31 is prime itself, so the answer is 31.

Add values of c such that when the graph is moved up, it has 4 intersections with the x-axis. 

This question seems to be a repost. I answered this already.

We are solving:

\(x(1.5) = 25 \quad \rightarrow x = \frac{25}{1.5} = \boxed{\frac{50}{3}}\)

\(x(2.5) = 42 \quad \rightarrow \quad x=\frac{42}{2.5} = \boxed{16.8}\)

\(87+213123+898 \cdot 1231231238- \frac{51235213}{5424554}\)

\((213210) + (1105645651724) - (9.445) = \boxed{1,105,645,864,924.555} \)

Let S be the weight of a small box and L be the weight of a large box. We can write a system of equations. Hope this helps.