Melody

I don't remember ever seeing a pronghorn before.  Thanks khalls2 and Brittany.

$$\\(x+2y)^3\\\\
=^3C_0(2y)^0(x)^3\;+\;^3C_1(2y)^1(x)^2\;+\;^3C_2(2y)^2(x)^1\;+\;^3C_3(2y)^3(x)^0$$

Can you take it from there?

That's great happy!

Perhaps anon wanted to know how to put a fraction into the calc or perhaps onto the forum.

It is hard to know what they really wanted.

So you did great   

Thanks Heureka.

That makes sense :)

X=1/3(90-x)

3X=90-X

4X=90

X=90/4

X=22.5

1/5(x+3)−2=18

1/5(x+3)=20

x+3=100

x=97

The graphical soln shows the real solution to be approx x=0.316

Here is the graph

4++2=4+2=6

Maybe you were supposed to ESTIMATE the answer

A solution of 63% fertilizer is to be mixed with a solution of 43% fertilizer to form 40 liters of a 53% solution. How much of the 63% solution must she use?

Choices     20L, 34L, 15L, 32L

The 63% solution is too strong by 10% and the 43% is too week by 10% so if i had to guess I would say 50% of each is needed.  That is 20L each.       

Again: This answer has  been edited.

Let X be the amount of 63% soln used and let Y be the amount of 43% solution used.

$$\begin{array}{rlll}
X+Y&=&40\qquad \qquad & (1)\\
0.63X+0.43Y&=&0.53(X+Y)& (2)\\
0.20X+(0.43X+0.43Y)&=&0.53*40\qquad $Error fixed - thanks Alan$\\
0.20X+0.43(X+Y)&=&21.2\\
0.20X+0.43*40&=&21.2\\
0.20X+17.2&=&21.2\\
0.20X&=&4\\
X&=&4/0.20\\
X&=&20\\
\end{array}$$

so it seems that 20L of the 63% solution was needed.