Melody

well say it had 5 pages,  1,2,3,4,5   You can see that the middle page is 3

Say it has got 6 pages   1,2,3,4,5,6    Then it has got 2 middle pages  3 ans 4

So the middle page will be the (total number of pages+1)/2

So you tell me Tangrachel what is the middle page/s?

Di you mean $$2x+\frac{1}{3}$$    

or do you mean   $$\frac{2x+1}{3}$$    ?   If this is what you mean then you need to use brackets.

sorry no fraction key

2 thirds would go in as 2÷3

5 and 3/4 take away 9 and 7/8

(5+3÷4)-(9+7÷8)=

I hope that helps. 

$$\left({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right){\mathtt{\,-\,}}\left({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{7}}}{{\mathtt{8}}}}\right) = -{\mathtt{4.125}}$$

Excellent answer anonymous!

Why don't you join up and then you could get proper credit! 

mean means average.

so add them up and divide by how many there are. 

This is your question

Sin ^2 63 + sin ^ 2 27/ cos ^17 + cos^ 73

This is what I think you actually meant

(Sin ^2 (63) + sin ^ 2 (27)/ (cos ^2 (17) + cos^2 (73))

You need to be more careful - anyway

$$sin\theta = cos(90-\theta)$$   and vise versa - you can see this if you draw a right angled triangle and take a look

therefore

sin63=cos27  and cos73=sin17    so the question becomes

$$\frac{cos^2 (27)+sin^2(27)}{cos^2(17)+sin^2(17)}\\
=\frac{1}{1}\\
=1$$

 Now please say thank you.  

1 16/16 = 1+1 = 2

But 1/4 of an inch is 4/16 of an inch and 13/16 is closer to 12/16 than it is to 16/16.

12/16 = 3/4

The answer is 1 3/4 inches  

Thanks for giving it a go anonymous!  We like everyone to join in.    

Good answer anonymous!

I just want to go over what you were saying.

Anything divided by itself is 1

so

 $${\frac{{\mathtt{7}}}{{\mathtt{7}}}} = {\mathtt{1}}$$

$$\frac{9^3}{9^3}=1$$

also

 $$\frac{5^7}{5^4}=5^7\div 5^4=5^{7-4}=5^3$$

so

$$\frac{9^3}{9^3}=9^{3-3}=9^0$$   BUT we know that    $$\frac{9^3}{9^3}=1$$

therefore

$$9^0=1$$      

Anything (except for 0) raised to the power of 0 is 1.     

$${{\mathtt{0}}}^{{\mathtt{0}}} = \underset{{\tiny{\text{Error: indeterminante}}}}{{{\mathtt{0}}}^{{\mathtt{0}}}}$$    Like it says 0⁰ is undefined    

Are you happy now bioschip? 

A really good answer Bioschip! 

Thanks Alan,

That is assuming you are talking about perpendicular height.