Alrighty
squaer base so length=width, nice
v=lwh
but in this case, l=w, so replace l with w
V=w²h
and volume is 32000
32000=w²h
the amount of materials is the surface area
note that there is no top
so
SA=LW+2H(L+W)
L=W so
SA=W²+2H(2W)
SA=W²+4HW
alrighty
we gots
SA=W²+4HW and
32000=W²H
we want to minimize the square foottage
get rid of one of the variables
32000=W²H
solve for H
32000/W²=H
subsitute
SA=W²+4WH
SA=W²+4W(32000/W²)
SA=W²+128000/W
take derivitive to find the minimum
dSA/dW=2W-128000/W²
where does it equal 0?
0=2W-1280000/W²
128000/W²=2W
128000=2W³
64000=W³
40=W
so sub back
32000/W²=H
32000/(40)²=H
32000/(1600)=H
20=H
the box is 20cm height and the width and length are 40cm
Answer: D
Step-by-step explanation:
10 squared is 100 and 11 squared is 121, and 111 is in between those two numbers.
Answer:
3 games
Step-by-step explanation:
The computation of the maximum number of games that he could buy is shown below:
= Amount available - already spent
= $75 - $28
= $47
If each game cost $15
So, the maximum no of games is
= $47 ÷ $15
= 3.133
= 3 games
Suppose that the farmer had bought the rice at x dollars per bag and had sold them at a 25% markup. How much did the bags cost him before he added the markup? 1.25x =$75 results in $75/1.25, or $60 per bag.
If he sold 25 bags, his profit would be 1.25($60/bag)(25 bags) = $1875.
I very seriously doubt that the rice was $7500 per bag. Perhaps you meant $75/bag...?
Answer:
x=1.2
Step-by-step explanation:
Ⓗⓘ ⓣⓗⓔⓡⓔ
1/2x-4=-3-1/3x
5/6x-4=-3
5/6x=1
x=1.2
(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥
