All Answers 358090 Answers

$$(x^{-2})^{-2}\div (x^{-2}-y^{-2})^{-2}\\\\
=(x^{-2*-2})\times (x^{-2}-y^{-2})^{+2}\\\\
=(x^{4})\times (\frac{1}{x^2}-\frac{1}{y^2})^{+2}\\\\
=(x^{4})\times (\frac{y^2-x^2}{x^2y^2})^2\\\\
=(x^{4})\times \frac{(y^2-x^2)^2}{x^4y^4}\\\\
=\frac{(y^2-x^2)^2}{y^4}\\\\
=\frac{y^4-2x^2y^2+x^4}{y^2}\\\\

.$$

No neither do I  

I accessed your graph just fine.  Although I did have to scroll in.  Thanks Chris  

$$-\sqrt{72}=-\sqrt{36*2}=-6\sqrt{2}$$

This does not translate properly.

I have tried using Google and Bing.

Perhaps you could check your spelling and try again.   

Have a  look here Dragon.

There is no working.  This is a super easy question Dragon.

|2|=2 

because 2 is 2 units from 0 on the number line. 

The 2 vertical lines are symbols for absolute value.

You need an = something in there 

Hi Sasini,

EDIT: SORRY I JUST READ WHAT ALAN WROTE.  HE HAS SHOWN THAT THEY ARE NOT EQUAL.

SO YOU WILL HAVE A TOUGH TIME PROVING THAT THEY ARE.  LOL.       (Thanks Alan)

I'D STILL LIKE YOU TO FOLLOW MY ADVICE FOR NEXT TIME  

We have done a couple of these for you Sasini.

Here is one we did a day or 2 ago.

$$\begin{array}{rlll}
2y-5\times 4&=&44 \qquad &\mbox{First you simplify the sides seperately }\\
2y-20&=&44 \qquad &\mbox{next you will need to get rid of the 20 } \\
2y-20+20&=&44+20 \qquad &\mbox{Now simplify again }\\
2y&=&64 \qquad &\mbox{Next you need the y by itself so get rid of the 2 }\\
2y\div 2&=&64\div 2 \qquad &\mbox{Simplify again }\\
y&=&32 \qquad &\mbox{ }\\
\end{array}$$

Just remember: What ever you do to one side you must also do to the other side. the equation must stay  balanced. 

$$(-1^2+11^2)(11^2- -7^2)\\
=(-[1^2]+11^2)(11^2+7^2)\\
=(-1+121)(121+49)\\
=(120)(170)\\
=120*170\\$$

$${\mathtt{120}}{\mathtt{\,\times\,}}{\mathtt{170}} = {\mathtt{20\,400}}$$

Do you mean like 3 different ways to represent a ratio?

4 to 3

4:3

4/3

2y + 3z =

2(8) + 3(-12) =

16 - 36 =

-20

4x =

4(-3) =

-12

The left-hand side is NOT equal to the right-hand side in general.  For example, suppose A = 37°

$${\frac{{cot}{\left({\mathtt{37}}\right)}{\mathtt{\,\times\,}}{csc}{\left({\mathtt{37}}\right)}}{\left({cot}{\left({\mathtt{37}}\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{37}}^\circ\right)}\right)}} = {\mathtt{1.037\: \!348\: \!331\: \!178\: \!024\: \!9}}$$

$${\frac{\left({cot}{\left({\mathtt{37}}\right)}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{37}}^\circ\right)}\right)}{\left({cot}{\left({\mathtt{37}}\right)}{\mathtt{\,\times\,}}{csc}{\left({\mathtt{37}}\right)}\right)}} = {\mathtt{0.239\: \!633\: \!701\: \!060\: \!557\: \!7}}$$

"I think there is an error in one of the equations there in the book. "

No, I think the error in the sign is a bug in the website calculator.  Mathcad gets the sign the same as the text book.

first solve the left side hand . bt, it's similar to right side hand

Thanks Dragon,

I did see the  -1 but it didn't worry me.  

You can never please all the people all the time and people are entitled to their 'vote'.    

Besides, our scores are jumping around so much at the moment that they don't mean anything anyway. 

It's ok dr

Thank You

3 +13(-11d + 10)

3 -143d + 130

133 -143d

I think it is the answer

3+13(-11d+10) =

3 -143d + 130 =

133 - 143d

I can't understand the quesion. sry

Thank you!

Thanks!

You got it correct!

When I played it, it helped. Thanks!