<em>given,</em>
<em>measure of sides of the</em><em> </em><em>square = 12 cm</em>
<em>we know,</em><em> </em><em>diagonal</em>
<em>d = a√2</em>
<em>now,</em>
<em>d = 12√2</em>
<em>d </em><em>=</em><em> </em><em>1</em><em>6</em><em>.</em><em>9</em><em>7</em><em>0</em><em> </em>
<em>a/</em><em>q </em><em>we </em><em>have </em><em>to </em><em>round</em><em> </em><em>the </em><em>answer </em><em>to </em><em>the </em><em>nearest</em><em> </em><em>centimetres</em>
<em>so,</em><em> </em>
<em>it </em><em>is </em><em>1</em><em>7</em><em> </em><em>cm</em>
<em>hope </em><em>this</em><em> answer</em><em> helps</em><em> you</em><em> dear</em><em>!</em><em> </em><em>take </em><em>care!</em>
P is 48
Q is 53
R is 22
S is 24
I believe you work the problem out and put the answer down.
Do you remember when you were learning about the "Least Common
Multiple" of two numbers, and you were wondering why you had to learn
it, because you were so sure that you'd never need to use it ? Do you
remember that ?
Well, HERE's where you get to use it. This question is just looking for
the LCM of 6 and 15 . THAT's how many days it will be before you get
homework in both subjects on the same day again.
(It's not 90 days, and it's not 60 days.)
This would be perpendicular
