Step-by-step explanation:
yes I am how about you, I always get stuck on my assignments
Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.
2(x+y) = 300
x+y = 150
y = 150-x
A=x(150-x) <--(substitution)
The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150
So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.
A=75(150-75)
A=75*75
A=5625
So the maximum area that can be enclosed is 5625 square feet.
A) $15 ÷ 40% =
15×.4= 6
15+6= 21
B) $38 × .25=9.5
38-9.5=28.50
28.50×.06=1.71
28.50+1.17=30.21
I think this is how you do it. would wait for other answers as well to double check.
First find the volume of the box: 8•4•4.5=144 cubic inches.
8 of the 1/8 inch=1 of the 1 cubic inch
He could do 144 and 0
143 and 8
142 or 16
Or just keep following the pattern... Hope that helped
