VooFIX

Yea, you were correct, Cubed root 3 was odd, not neither. The answer was 1,3,4

ok thanks, i'll check

my bad i didn't realize i typed -133 instead of -113

Well, the Give Up button was way. Now I understand fully how to solve this question. ;-;

https://imgur.com/a/W7b2gG6 

This was supposed to be a question on my math hw-

I tried your answer and method. That's incorrect.

Justice for Floyd! 

Guest consider this, a palindrome is a number that is the same reversed. Eg. 1234567 is not a palindrome because 7654321 is not the same. 11111111 is because it is the same reversed. You want a 7 digit palindrome so you have 7 places to fill. 

__ __ __ __ __ __ __
9 x 10 x10x10x1x1x1

Numbers are options^

You cannot fill the 1st spot with 0 So the first and last digit will be eliminated of 0. So there is 9 digits from 1-9. If you choose 9 for the first one the last one must be the same. So the last one is 1 option. Once the options are filled multiple! 

\(9 * 10 * 10 *10*1*1*1 = 9000 \)

So there should be 9000 options for 7 digit palindromes! I hope thats correct!

Is there an image? Also is it possible if you can type that out of Latex as some of the information is not there. Very hard to read xd.

Its 36

Nope, I've tried but nope...

Thanks!

Check if 60 is correct

A rough sketch shows a square, of side L, with 5 rectangles inside, each L by L/5

Perimeter of rectangles = 2L +2(L/5) = 2.4 L = 36,,,,L= 15
Perimeter of square = 4*L=4*15 = 60 units

Is my answer correct? 60

Let x be number of total enrolled then 20% of the OD, out of district, students would be 0.2x. The number is increased by 50%, 1.5 so (0.2x)(1.5)=0.3x, so 30% of the total enrolled is out of district.

Wait since its isosceles then the base should be 13 and the hypotenuse is 14 and therefore we can solve the height. In Isosceles, the angle bisector is the perpendicular bisector, pretty sure. 

Hi, this forum is basically you post a question and get an explanation and answer. The moderator thing I'm not sure, you might need to get recommended. When a mod is online, send a private message or just create a new question asking about it.