CPhill

14  and 15

No real solutions.....two "non-real".....

$${\mathtt{0.5}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{1}}{i}{\mathtt{\,-\,}}{\mathtt{3}}\\
{\mathtt{x}} = {i}{\mathtt{\,-\,}}{\mathtt{3}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,-\,}}{i}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{i}\\
\end{array} \right\}$$

Each side of the cube has an area of s^2, where s is the side length.

And since there are 6 sides, the total surface area is just 6s^2.

The library has an area of 400 sq. ft....but......we can't say what the perimeter might be

The library could have  the dimensions  20 x 20, in which case the perimeter is 80 ft.

Or, it might have the dimensions  40 x 10, in which case the perimeter would be 100 ft.

In short, we would have to know the dimensions of the library to determine the correct perimeter.

Sorry, I don't understand this question........(????)

20.4%   = .204  = 204/1000 = 51 / 250

The sum, S_n,  is given by  :

(n/2)[(2a + (n-1)d ]    where a is the first term = 4,  n = 126  = the number of terms,   and d = -3, the common difference

So we have

(126/2) [ 2(4) + (125)(-3)] =

-23121

The new side length  = √90.25  = 9.5 inches

If the original width was 24 inches  and the logo is 6 inches, than it appears that the logo has a scale factor of 1/4 of that of the picture. So, the logo height  = (1/4)* 36   = 9 inches