All Answers 355940 Answers

You're probably young-ish enough to be my granddaughter!  THAT's how old I am.... LOL   

Please help

Correct 

Thank you I just reported it! I hope it works 

I am sorry that you are having this problem.

I am not involved with the technical side of the forum at all.

I  will try to get input from admin for you but maybe just try again every few days and see if it works.

I am sorry that I cannot be more helpful.

Actually,  You can lodge a complaint, which admin will probably see, and then he will have your email address because I think you will have to include it

There is a link at the bottom of the page saying 'report a problem'  try doing that. 

I hope this is helpful.  Get back to me with an update.  I hope you can get your membership to activate.    

Melody Can you help me?

When I try to register it wont send me an email in Inbox or in spam and my email is correct.

Do you know why it won't send an email?

famous singer or famous person

:THIS IS A SCREENSHOT ⬆️⬆️⬆️⬆️

And Sorry I don't know the perimeter of this net

alot...

10080 , 12600 , 15120 , 17640 , 20160 , 22680 , 25200 , 27720 , 30240 , 32760 , 35280 , 37800 , 40320 , 42840 , 45360 , 47880 , 50400 , 52920 , 55440 , 57960 , 60480 , 63000 , 65520 , 68040 , 70560 , 73080 , 75600 , 78120 , 80640 , 83160 , 85680 , 88200 , 90720 , 93240 , 95760 , 98280 , Total =  36 such numbers.

@Guest, I hope it helped!

You just helped me :)

They work

the second one is the proper link.

I like your song.

What do you want help with?

18/23

Dear guest,

I would suggest you not use terms such as 'asap' in your questions.

It implies that you think your question is more important than other people's.

Or worse still that you have left your homework till the last minute, or you are doing a test and hence late help is equal no help at all.

Terms such as this, make me less inclined to respond quickly.

= \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }

\(z= \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }\\ z= \frac{ 3 - 2 *2 }{\sqrt{3} - 2\sqrt{2} }\\ z= \frac{ -1 }{\sqrt{3} - 2\sqrt{2} }\\ z= \frac{ -1 }{\sqrt{3} - 2\sqrt{2} }*\frac{\sqrt{3} + 2\sqrt{2} }{\sqrt{3} + 2\sqrt{2} }\\ z= \frac{-1(\sqrt{3} + 2\sqrt{2}) }{3 - 8}\\ z= \frac{(\sqrt{3} + 2\sqrt{2}) }{5}\\ z= \frac{(almost 2 + almost3)}{5}\\ z= \frac{almost 5 }{5}\\0

This would not display before so here is a pic of my answer.

if z \(= \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }\)  find  \(\lfloor z\rfloor \)

Hello Guest!

\(z= \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }\)

\(z=\dfrac{(\sqrt{3}+\sqrt{2}\cdot \sqrt{2})(\sqrt{3}-\sqrt{2}\cdot \sqrt{2})}{\sqrt{3}-2\cdot \sqrt{2} }\\ z=\dfrac{(\sqrt{3}+\sqrt{4})(\sqrt{3}-\sqrt{4})}{\sqrt{3}-2\cdot \sqrt{2}}\\ z=\dfrac{3-4}{\sqrt{3}-2\cdot \sqrt{2}}\\ \color{blue}z=\dfrac{1}{2\cdot \sqrt{2}-\sqrt{3}}=0.9121\)

\(\color{blue}\lfloor z\rfloor = 0\)

Thanks to Alan.

  ! 

Thanks!!

If the area of a parallelogram is increased by a factor of 25, assuming both dimensions change proportionally, what is the effect on the base and height?

The  base  and  height  change  by  a  factor of  sqrt (25)  =   5

The radius of a circle is multiplied by 2/3, describe the effect on the circumference?

Circumference is a  linear measure......it will  also  be (2/3)  of its  original value