SA=2(LW+LH+WH)
SA=94
L=3
W=4
94=2(3*4+3H+4H)
divide both sides by 2
47=3*4+3H+4H
47=12+7H
minus 12 both sides
35=7H
divide oth sides by 7
5=H
5m
(x+2)(4+y)(5+6)(2+x)(4+0.5x)
combine like terms
(x+2) (4+y) (11) (x+2) (4+.5x)
11(x+2)^2 (4+y)(4+.5x)
distribute
11(x^2 +4x+2) (16+4y+2x+.5xy)
(11x^2+44x+22)(16+4y+2x+.5xy)
5.5 x^3 y + 22 x^3 + 66 x^2 y + 264 x^2 + 198 x y + 792 x + 176 y + 704
We can add 1/6 & 1/3 together to find out the total amount of crackers that were eaten:
1/6 + 1/3 (Find the Common Denominator. In this case it is 6.)
= 1/6 + 2/6
= 3/6 (Now find the Lowest Terms. Both 6 & 3 divide into 3)
= 1/2 (In Lowest Terms)
This means that 1/2 of the Crackers have been eaten.
1/2 of 18 = 9
Therefore, Bob has 9 Crackers left.