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well, y must be greater than x, so x + 2 = y.

$y^2-x^2=384$ and $x + 2 = y$

so solving that system of equations... u get $x=95, \: y=97$

so 97 + 2 = 99

Hi there, 

i cant see ur steps but here are mine:

$\left(x+2\right)\left(2x-3\right)\le-3$

$2x^2+x-3\le0$

$\left(x-1\right)\left(2x+3\right)\le0$           (u get this by factoring $(x-1)(2x+3)$)

$-\frac{3}{2}\le x\le 1$                          (identify the interval)

oh i just noticed my message didn't post! yes i got it

i've done this one, u did it correct!

omg how did i not see this! Here is how u can solve the system:

1. Isolate $l$ in $2(l+w)=23$ and get $l=\frac{23}{2}-w$
2. Plug the solutions $l=\frac{23}{2}-w$ into $\left(l+1+1\right)\left(w+1+1\right)=58.5$
3. For $\left(l+1+1\right)\left(w+1+1\right)=58.5$ subsitute $l$ with $\frac{23}{2}-w: \:\:w=\frac{9}{2},l=7$

yeah i got it

HINT:

well, there are 25 multiples of 4.

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100

Solving this equation, x can be 3 or -1.

3² = 9 and (-1)² = 1

9 + 1 = 10