XxmathguyxX

I got a different answer from Cphill...

We know sum of all the interior angles of a polygon having n sides = (n-2)180

Given smallest angle = 120

a = 120

d = 5

Sum, S = (n/2)(2a+(n-1)d)

(n/2)(240+(n-1)5) = (n-2)180

n(240+5n-5) = (n-2)360

5n2+235n-360n+720 = 0

5n2-125n+720 = 0

n2-25n+144 = 0

(n-9)(n-16) = 0

n = 9 or 16.

n = 16 cannot be possible since interior angle cannot be greater than 180.

I think you will find this useful:

yeah...

Ok, so if there is a value to b, it would be

\(20^2+b^2=c^2\)

\(400+b^2=c^2\)

so yeah...

ok,

According to the Pythagorean Thereom,

\(12^2+b^2=20^2\)

\(144+b^2=400\)

\(b^2=256\)

\(b=16\)

This is for 19.

 \(25+100=125\)

\(\sqrt125=15\)

so 15 for number 20

odd*odd*odd =odd 

which is 1/2*1/2*1/2=1/8 chance of odd, so 7/8 chance of even

so \(\frac{2}{8}=66\)

\(\frac{1}{8}=33\)

\(\frac{3}{8}=99\)

so she saved 99 dollars!

(so she saved sounds like a tounge twister to me!)

\(9^\frac{1}{2}\times9^\frac{1}{2}=\)

\(3^2(\frac{1}{2})\times3^2(\frac{1}{2})\)

 it is 3^2x1/2, not 3^2 and then x1/2

\(3\times3\)

\(9\)

@ guest I think you are wrong

reason:

'Dylan's dimes total $1.30 less than his quarters "

so the quarters have to be at least 6 of them

My attempt:

So we need 2x as many nickels as dimes

so

Thhis is acually a pretty bad method of doing it but if you noticed, there is a pattern

I'll let you find it

okay,

\(x^2+y^2=185\)

\(x+55=2y\)

\((2y-55)^2+y^2=185\)

\(4y-110+y^2=185\)

\(4y+y^2=295\)

\(y^3=\frac{295}{4}\)

wait wait wait, you said x and y were integers....

um.. Melody or Cphil plz help?

ps. 8 and 11 works for the second statement

pps. 13 and 4 also work for the second statement

I'm not sure if this problem is possible since x and y are whold numbers....

My shot: 

\(2a+2b+2c=b\)

\(a+b+c=\frac{b}{2}\)

\(-\frac{b}{2}=-5+b\)

\(-5=-\frac{3b}{2}\)

\(-10=3b\)

\(b=-\frac{10}{3}\)

\(a=\frac{34}{3}\)

\(c=-\frac{59}{3}\)

\(a\times b\times c=\frac{20060}{9}\)

huh, straight we got different answers......