mworkhard222

let x = # of red and blue stickers

2x = total stickers

\(2x-\frac23x-125=\frac14\cdot2x\)

\(2x-\frac{2}{3}x=\frac{1}{2}x+125\)

\(\frac{5}{6}x=125\)

\(5x=750\)

\(x=150\)

Question #1: \(\frac{125}{150}=\frac{5}{6}\)

Question #2: 150

1.

     a) 7

     b) 11

2. √250 \(\approx\) 15.81
    15.81 * 4 = 63.24
    So 63

sure, you said i could use a calculator so i did, then i rounded it to the nearest hundredth

a) 4.47
b) 5.92
c) 10.72

a monomial i think

ok, not thay hard. we can make a variable.

say x = pieces of candy Sam has.

so Kabir has 5x

and the total is x + 5x = 6x

then we know that 5x - 18 = 3x        (the 3x is from 6x/2)

\(5x-18=3x \)

\(5x=3x+18\)

\(2x=18\)

\(x=9\)

so the total is 6x, 6 * 9 = 54.

then that doesn't exist

if x and y were the side lengths, then, \(2x+y=30\)

Here are all the lengths but we'll take out the ones that don't satisfy the triangle inequality:

1,1,28

2,2,26

3,3,24

4,4,22

5,5,20

6,6,18

7,7,16

8,8,14

9,9,12

10,10,10

11,11,8

12,12,6

13,13,4

14,14,2

Final answers:

8,8,14

9,9,12

10,10,10

11,11,8

12,12,6

13,13,4

14,14,2

 7 of them

Lets simplify the equation first:

\(x^2+\frac{1}{x^2}\)

\(=\frac{x^4+1}{x^2}\)

so, 

\(\frac{23}{x^2}\) is the answer