1.11^5   OR     1.11 x^y 5

This is what you enter into many calculators

The web2.0 calc is 1.11^5

$${{\mathtt{1.11}}}^{{\mathtt{5}}} = {\mathtt{1.685\: \!058\: \!155\: \!1}}$$

You can find the web2.0 calc on the home page.

Look at it like this

If you have $10 and you spend $6 you will have $4 left RIGHT? WHY IS THAT?

It is because 10-6=4

Now if the numbers are big or if letters are used instead of numbers, then you still do the same thing!

If you have $10 and you spend x dollars then you will have 10-x   dollars left.     (see i swapped the 6 for the x)

--------------------

Now your problem

Kala earns $701 and she spends d dollars how much will she have left?

Copy the example above and you tell me what the answer is.    

A LaTex question. Heureka could you help me please.   

I just split that answer into 2 arrays so that I could put the full line comment in.

For this answer that is fine but sometimes it would mean that the equal signs will not line up.

Do you know how to do this without starting a new array?

Thank you.

Melody.

$$\begin{array}{rll}
\frac{5y^2}{y^2-16}&=&\frac{10}{y-4}+\frac{6y}{y+4}\\\\
\frac{5y^2}{(y-4)(y+4)}&=&\frac{10}{y-4}+\frac{6y}{y+4}\qquad

\mbox{You should notice that the left denomonator is the diff of 2
squares}\\\\
\end{array}
\\\mbox{The easiest thing to do is to multiply both sides by the lowest common denominator which is (x-4)(x+4)}\\\\
\begin{array}{rll}
\frac{5y^2(y-4)(y+4)}{(y-4)(y+4)}&=&\frac{10(y-4)(y+4)}{y-4}+\frac{6y(y-4)(y+4)}{y+4}\qquad \\\\
5y^2&=&10(y+4)+6y(y-4)\qquad \\\\

\end{array}$$

And you should be able to finish it from there.

If you don't understand or you need more help let us know what your problem is and we will help you more. 

Yes do that, Ninja is a very good teacher!

basically you have said

12*d = d

12*d = 1*d

d is dollars so you have said that if I buy one it is the same total price as if I buy 12.  That is not right!

1 pen will cost $d    Cost=$d

2 pens will cost $d*2 = 2d dollars

3 pens will cost 3d dollars

--

12 pens will cost 12d dollars

n pens would cost nd dollars

I think the answer that your teacher wants is just

cost of 12 pens = 12d dollars

or just

Cost=12d    dollars

|14-15|-|-11|

=|-1|-+11

=1-11

=-10

If you want me to explain any of this better then just ask.  

  (2cos 40 - cos 20)/sin 20

$$[2cos^220-2sin^220-cos20]/sin20\\

=[2-2sin^220-2sin^20-cos20]/sin20\\

=[2-4sin^220-cos20]/sin20\\

=2cosec20-4sin20-cot20\\

=\mbox{I am just playing here, Aziz is right, there is not much you can do with this}$$

You can enter it into the web2 calc like this.  

 (2*cos(40)-cos(20))/sin(20)

=$${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{40}}^\circ\right)}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{20}}^\circ\right)}\right)}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{20}}^\circ\right)}}} = {\mathtt{1.732\: \!050\: \!807\: \!567\: \!060\: \!3}}$$

Okay the answers are different.  

Aziz has assumed radians, which is a correct assumption because it you do not specifiy the units then by default it is radians.

I have used degrees because I am used to school students and I assume that is what you actually wanted.

So 2 different interpretations (both reasonable) have produced 2 different answers.

I have done this with a binomial expansion which you may not have done yet.

I have taken more steps than necessary to make it easier to follow.

A pop up has shown me that AziaHusain has done it by  a normal expansion.  

Hopefully our answers are the same.   

$$(a^2+3b)^3\\\\
=^3C_0(a^2)^3(3b)^0+^3C_1(a^2)^2(3b)^1+^3C_2(a^2)^1(3b)^2+^3C_3(a^2)^0(3b)^3\\\\
=(a^2)^3+3(a^2)^2(3b)^1+3(a^2)^1(3b)^2+(3b)^3\\\\
=a^6+3a^4*3b+3a^2(3b)^2+(3b)^3\\\\
=a^6+9a^4b+3a^2(9b^2)+27b^3\\\\
=a^6+9a^4b+27a^2b^2+27b^3\\\\$$

(4/52) * (3/51)*(48/50)*3

That's my first guess.

Sounds good to me. :)