All Answers 358087 Answers

Why is that true Alan?

Is there something obvious that I am missing?

Use the chain rule to get:

$$\frac{d\ln{p}}{dt}= \frac{d\ln{p}}{dp}\frac{dp}{dt}=\frac{1}{p}\frac{dp}{dt}$$

Can find this from:

$$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}$$

$${\frac{{\left({\mathtt{2\,014}}{\mathtt{\,\times\,}}{\mathtt{2\,015}}\right)}^{{\mathtt{2}}}}{{\mathtt{4}}}} = {\mathtt{4\,117\,267\,101\,025}}$$

verrrry niiiice ... 

I understand all question 

thank you melody and aziz

$${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{105}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{\mathtt{34}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{8\,849}}}}{\mathtt{\,-\,}}{\mathtt{105}}\right)}{{\mathtt{32}}}}\\
{\mathtt{t}} = {\frac{\left({\sqrt{{\mathtt{8\,849}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{105}}\right)}{{\mathtt{32}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{0.341\: \!589\: \!889\: \!970\: \!270\: \!7}}\\
{\mathtt{t}} = {\mathtt{6.220\: \!910\: \!110\: \!029\: \!729\: \!3}}\\
\end{array} \right\}$$

If you want to do it by hand use the quadratic formula 

a=16, b=-105 and c=34

If you do not know the quadratic formula, here it is.

nvm i think the answer is too big its impossible to calculate

X 10^(-5) means move the decimal place to the left 5 times.

7.5 --> 0.000075.

Question 10: Note that since the lines are parallel, their slopes must be the same [and negative since the line is going downward] and they also differ in their y-intercepts [one + and one -] as well based on the graph. 

D does have the same slope, but notice that both y-intercepts are positive...the graph has 1 equation that has a negative y-intercept.

C also has one that is x -y which means the x would become positive again once it is moved to the other side...while the other equation still has a negative x...

A has x, 2x, and 2y, so notice that the slopes would dramatically differ...

Only B makes sense since you notice that x + y = # andx + y = -#...the slopes are equal and the #'s are opposite signs.

Question 16: y is definitely greater than 0...but notice that is restricted to a maximum of y = 3. Only choice C makes sense.

Question 19: On the graph, when x = 1 when y = 0.

y = -ln x --> 0 = - ln(1)...Yes!

y = e^-x --> 0 = e^(-1)...No!

y = e^x --> 0 = e^1...No!

y = ln x --> 0 = ln 1...Yes!

Now, notice that if we took ln(0.9) we would get -0.1053605156578263...and this graph is in the positive range when x <1...so, -lnx would make it positive...A.

I'm going to do the last one  

$$y=\sqrt{25-x^2}\\
when\; x=4\\
y=\sqrt{25-16}=\sqrt9=3$$

So you have a rectangle with length 2 units and height 3 units.

So the area is 2*3=6u^2

Interesting...I wonder what type of questions appear on the NY & CA Common Core! I remember taking the CSTs (California Standardized Test) a long time ago and those questions were pretty hard at the time...

Yes, I forgot about the left-to-right rule...sorry about that. 

Esta pregunta necesita numeros y mas informacion!

You can think of this as moving the decimal place 34 times to the right...a general rule for anything that is in a similar format, X 10^(#), is you move the decimal place either to the left, if the exponent is negative, or to the right, if the exponent is positive. In this case, we have 34, a positive number, so we have to move the decimal place 34 times to the right and continue to add zeros until we reach the desired position.

0.20 --> 2000000000000000000000000000000000.

You can probably do this on the web2 calc as well but i don't know how.

anyway, I got you the answer. 

This question doesn't make sense. 

I didn't know that Chris.  Thanks 

[ x/(3+x) * (2+x)/(3+x) ] / (1-x)/(3+x) = 0,2336

$$[ x/(3+x) * (2+x)/(3+x) ] / (1-x)/(3+x) = 0,2336\\\\
\left[\frac{x}{(3+x)} \times \frac{(2+x)}{(3+x)}\right]\div \frac{(1-x)}{(3+x)} = 0.2336\\\\
\mbox{It should be noted straight off that }3+x\ne0\;so\;x\ne-3\\\\
\left[\frac{x}{(3+x)} \times \frac{(2+x)}{(3+x)}\right]\times \frac{(3+x)}{(1-x)} = 0.2336\qquad \mbox{the (x+3) cancels,and }x\ne1\\\\
\left[\frac{x}{(3+x)} \times \frac{(2+x)}{1}\right]\times \frac{1}{(1-x)} = 0.2336\qquad \mbox{}\\\\
\frac{x(2+x)}{(3+x)(1-x)} = 0.2336\qquad \mbox{}\\\\
\frac{x(2+x)}{(3+x)(1-x)} = 0.2336\qquad \\\\
\mbox{Now multiply everything by the lowest common denominator (3+x)(1-x) to get rid of the fraction}\\\\
x(2+x) = 0.2336(3+x)(1-x)\qquad \\\\$$

you can keep going from here and solve it with the quadratic formula or you can just plug your initial equation into the site calc and get it to do it all for you.  

(x/(3+x) * (2+x)/(3+x) ) / (1-x)/(3+x) = 0.2336  

change the square brackets to round ones and change your decimal comma to a decimal point and just plug it in.

$${\frac{{\frac{\left({\frac{{\frac{{\mathtt{x}}}{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}}{\mathtt{\,\times\,}}\left({\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}}\right)}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{x}}\right)}}}{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}} = {\mathtt{0.233\: \!6}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{\,-\,}}{\frac{{\mathtt{14\,508}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}}{{\sqrt{{\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{219\,876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\mathtt{704\,375}}}}}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{704\,375}}}{\left({\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{502}}}{{\mathtt{219}}}}}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{219\,876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\mathtt{704\,375}}}}}{\left({\mathtt{876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}\right)}}{\mathtt{\,-\,}}{\mathtt{2}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{\,-\,}}{\frac{{\mathtt{14\,508}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}}{{\sqrt{{\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{219\,876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\mathtt{704\,375}}}}}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{704\,375}}}{\left({\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{502}}}{{\mathtt{219}}}}}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{219\,876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\mathtt{704\,375}}}}}{\left({\mathtt{876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}\right)}}{\mathtt{\,-\,}}{\mathtt{2}}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\frac{{\mathtt{14\,508}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}}{{\sqrt{{\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{219\,876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\mathtt{704\,375}}}}}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{704\,375}}}{\left({\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{502}}}{{\mathtt{219}}}}}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\sqrt{{\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{219\,876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\mathtt{704\,375}}}}}{\left({\mathtt{876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}\right)}}{\mathtt{\,-\,}}{\mathtt{2}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\frac{{\mathtt{14\,508}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}}{{\sqrt{{\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{219\,876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\mathtt{704\,375}}}}}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{704\,375}}}{\left({\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{502}}}{{\mathtt{219}}}}}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\sqrt{{\mathtt{191\,844}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{219\,876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\mathtt{\,-\,}}{\mathtt{704\,375}}}}}{\left({\mathtt{876}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{625}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{6\,484\,410\,131}}}}}{\left({\mathtt{219}}{\mathtt{\,\times\,}}{{\mathtt{146}}}^{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{10\,930\,245\,625}}}{{\mathtt{84\,027\,672}}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{6}}}}\right)}\right)}}{\mathtt{\,-\,}}{\mathtt{2}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{3.318\: \!755\: \!885\: \!933\: \!879\: \!7}}{\mathtt{\,-\,}}{\mathtt{2.004\: \!784\: \!814\: \!651\: \!891\: \!5}}{i}\\
{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{3.318\: \!755\: \!885\: \!933\: \!879\: \!7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2.004\: \!784\: \!814\: \!651\: \!891\: \!5}}{i}\\
{\mathtt{x}} = -{\mathtt{2.184\: \!609\: \!027\: \!146\: \!14}}\\
{\mathtt{x}} = {\mathtt{0.822\: \!120\: \!799\: \!013\: \!899\: \!5}}\\
\end{array} \right\}$$

My goodness that looks horrible!

Let's see what the graph looks like and get the answer/s from there.

Alright there are your 2 answers.  I might keep going with the quadratic formula and get them that way.

$$x(2+x) = 0.2336(3+x)(1-x)\qquad \\\\
x^2+2x = 0.2336(3-3x+x-x^2)\qquad \\\\
x^2+2x = 0.2336(-x^2-2x+3)\qquad \\\\$$

I could just finish this by hand but I'mm goint to let the web 2 calc do it for me.

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}} = {\mathtt{0.233\: \!6}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right) \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{103\,571}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{257}}\right)}{{\mathtt{257}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{103\,571}}}}{\mathtt{\,-\,}}{\mathtt{257}}\right)}{{\mathtt{257}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{2.252\: \!235\: \!355\: \!360\: \!168\: \!6}}\\
{\mathtt{x}} = {\mathtt{0.252\: \!235\: \!355\: \!360\: \!168\: \!6}}\\
\end{array} \right\}$$

There you go, that is better!

These 2 answers would have been included in the original calculator ouput but some other imaginary roots were included as well.  You probably do not need to worry about those. 

Hi rdplymale,

To start with h does not equal 2. h does not equal anything because there is not h in the equation.

There are many ways of looking at this, I will examine 1 and come back and look at others if you wish.

Yes the axis of symmetry is x=2

Since this is a variation on y=x^2 it is a concave up parabola.  Do you already understand this?

The axis of symmetry of y=x^2 is x=0

Since it is (x-2)^2 the axis of symmetry is moved so that x-2=0  that is x=2

The minus 5 just moves every point down by 5.

I'll demonstrate with some graphs.  

Let's divide and see..

17 1/2 = 35/2

So

(35/2) / (1/6) =

(35/2) * (6/1) =

35 * 3  =

105

And that's the answer

The geometric mean is defined as the nth root of the product of n numbers.......

So we have

  ⁵√(110*50*80*120*90) =  ⁵√(4,752,000,000) ≈ 86.17

I just realised that you have an n and a k in the same equation so I won't be able to solve that one anyway.

How about I make up one of my own.

$$\frac{2x}{3}-\frac{5}{6}=\frac{5}{2}$\\
The lowest common denominator is 6 so multiply both sides by 6\\
6\times\left(\frac{2x}{3}-\frac{5}{6}\right)=6\times{\frac{5}{2}}\\\\
\frac{6\times 2x}{3}-\frac{6 \times 5}{6}=\frac{6 \times 5}{2}}\\\\
\frac{2\times 2x}{1}-\frac{1 \times 5}{1}=\frac{3\times 5}{1}}\\\\
4x-5=15\\
4x=20\\
x=5\\$$

Of course it really depends on the exact problems but this is a method that I use a lot of the time. 

Sun 10-8-14

* Complicated 'quadratic' equation - extension question, for interest only.

@@ End of Day Wrap :      Sat 9/8/14      Sydney, Australia     Time 2:45  am    (Really Sunday morning)

Hi everyone,

Are you all enjoying your  weekend?  I hope that you are.  

There sere soem great answers today from CPhill, Rosala, alan, DragonSlayer554, AzizHusain, and NinjaDevo.

A big thank you to each of you.    

Rosala is  back, she has been away because of illness.  We are all glad that you are all better now. 

Here are the interest posts for the day:

(1) Rosala's back 

No DragonSlayer , that isnt also how it works!

having you points above 1000 will give u a star or make u a power user but making u a moderator can only be done by the site's owner that is Andre Massow!

if he felt u are fit to be a moderator he would make u one as he made Radix the moderator of the German forum!

But for you DragonSlayer , i dont think Andre wil make u a moderator becoz u are too young to be a moderator , and not only you infact me , Zegroes, reinout and all!and about the rest , i dont know if they r grown up or small!

so i hope uve understood it b now!

Thank you :D