CPhill

9P^2 - 16Q^2   this is a difference of two squares...to factor it, take the square root of both terms

3P and 4Q

Write them once separated by a plus and once separated by a minus

(3p + 4Q) (3P - 4Q)      and there you go

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Suppose we had

30 = 2P + 5Q     to write this in terms of P, subtract 5Q from both sides

30 - 5Q  = 2P   now, we just want P by itself, so divide both sides by 2

[30 - 5Q] / 2   = P

And we've written this in terms of P......

Your answer isn't incorrect, saseflower.......you're just more "literary"....that's all.....

5[f(d)]..... f(x)=-2x-3

This says to put "d" into the function and then multiply the result by 5......so we have

-2(d) -3

5 [ -2(d) - 3 ] =

-10d - 15

Nice answer, saseflower.....

This is a parabola that opens upward.

The x coordinate of the vertex is given by -b/2a  = -2/(2*4) = -1/4

And putting this back in the original problem to find the y coordinate of the vertex, we have   y = 4(-1/4)^2+2(-1/4) - 7  =  4(1/16) - 1/2 - 7 =  1/4 - 1/2 - 7 = -1/4 - 7  = -29/4

The axis of symmetry is x = -1/4

The y-intercept is when x =0  = (0, -7)

We have a 1/2 chance of getting heads on any flip.......our task is to do this three times in a row.....so we have

(1/2)^3  =  1/8

Here's one more way.....for argument's sake let the original price be $100.

Then, the new price is just $60

So to find the percent increase to raise this back to the original price we have

[100- 60]/60 x 100%  = 40/60 x 100%  = 2/3 x 100%  = about [66 + 2/3] %

Here are the values....

$${\mathtt{\,-\,}}{\mathtt{9}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{133}}}}{\mathtt{\,-\,}}{\mathtt{5}}\right)}{{\mathtt{18}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{133}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}{{\mathtt{18}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.362\: \!920\: \!144\: \!148\: \!377\: \!5}}\\
{\mathtt{x}} = {\mathtt{0.918\: \!475\: \!699\: \!703\: \!933\: \!1}}\\
\end{array} \right\}$$

a+b=4ab   b+c=5bc   c+a=3ac

a= b(4a - 1)     a = c(3a - 1)

b = a/(4a -1)  c = a/(3a -1)

b + c = 5bc

a/(4a -1) + a/(3a -1)  = 5a^2/[(4a -1)(3a -1)]

[a(3a -1) + a(4a -1)]/ [(4a -1)(3a -1)]  = 5a^2/[(4a -1)(3a -1)]

3a^2 - a + 4a^2 - a  = 5a^2

2a^2 - 2a = 0

a^2 - a = 0

So

a(a -1) = 0     0 is trivial....so a = 1

So  b = a/[4(a) -1] = 1 / [4(1) -1] = 1/3 

And  c = a/[3(a) -1] = 1 /[3(1)-1] = 1/2

Check

a + b = 1 + 1/3 = 4/3 = 4ab =4(1)(1/3) = 4/3  

b + c = 1/2 + 1/3  = 5/6 =5bc = 5(1/3)(1/2) = 5/6

a + c = 1 + 1/2 = 3/2 = 3ac = 3(1)(1/2) = 3/2

In answering this question, I'm assuming that all the tiles used are "whole"...no "partial" tiles are used.

If the room is 16m^2, this means that each side of the room is 4m = 400cm. And if 256 tiles were used of type P, there must have been 16 on each side. Then each one would have been .25cm on each side. To see this, note that (.25cm)x16 = 400cm

Now, if whole tiles are used, and they have to be more than 15 cm on a side.....we need to find all the side lengths between 15 and 25 that divide 400 evenly. Only two fit the bill.....a side length of 16cm and a side lenth of 20cm.

If a side length of 16cm is used  (400/16)^2 = 625 tiles of type R must be used.

If a side length of 20cm is used (400/20)^2 = 400 tiles  of type R must be used.