SA=2H(L+W)+LW for open box
20=2H(4+W)+4(W)
distribute
20=8H+2WH+4W
divide both sides by 2
10=4H+WH+2W
solve for 1 variable, pick W
10-4H=WH+2W
10-4H=W(H+2)
(10-4H)/(H+2)=W
V=LWH
subsitute 4 for V, subsitute H for H and (10-4H)/(H+2) for W
V=(4)(H)(10-4H)/(H+2)
V=(40H-16H²)/(H+2)
find max value
take deritivitive of this thing
V'=-16(H²+4H-5)/((H+2)²)
using sign chart
sign changes from positive to negative at H=1
so at H=1
find W
W=(10-4H)/(H+2)
W=2
the dimeionts are
length=4ft
width=2ft
height=1ft
(the volume is 8 cubic feet)
The answer -35.5 Hope this helps you
Answer:
Step-by-step explanation:
y+5 = ¾(x+4)
Answer:D) 1 hr
Step-by-step explanation:
Given
Pump A and B can fill tank in 
Pump A and C can fill tank in 
Pump B and C can fill tank in 
Let A be the total hr taken A therefore rate of 
Let B be the total hr taken B therefore rate of 
Let C be the total hr taken C therefore rate of 
-----1
-----2
-----3
Adding 1,2 & 3
thus time taken by A,B and C combined is 1 hr
The answer is 3x-x :)
You just simplify the expression, (4x^2-x^2) - (x- (-2x))
