Let
x------> the length side of the square base of the box
y-------> the height of the box
we know that
volume of the box=b²*h
b=x
h=y
volume=256 cm³
so
256=x²*y------>y=256/x²--------> equation 1
<span>The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.</span>
surface area of the cardboard=area of the base+perimeter of base*height
area of the base=x²
perimeter of the base=4*x
height=y
surface area=x²+4x*y-----> equation 2
substitute equation 1 in equation 2
SA=x²+4x*[256/x²]-----> SA=x²+1024/x
step 1
find the first derivative of SA and equate to zero
2x+1024*(-1)/x²=0------> 2x=1024/x²----> x³=512--------> x=8 cm
y=256/x²------> y=256/8²-----> y=4 cm
the answer is
the length side of the square base of the box is 8 cm
the height of the box is 4 cm
Answer:
100
Step-by-step explanation:
Answer:
it has 11 faces 10 sides for the decagon + 1 for the base
It does have one face that is a decagon, which is the base, the rest are triangles
it has 20 edges
Step-by-step explanation:
Answer: One plain roll is 4 dollars and one shiny roll is 6 dollars.
Step-by-step explanation:
lets start by saying
rolls of plain wrapping paper = x
rolls of shiny wrapping paper = y
Kathryn sold 4 plain rolls and 3 shiny rolls for 34 dollars.
4x+3y=34
Eugene sold 4 plain rolls and 2 shiny rolls for 28 dollars
4x+2y=28
Both equations will look like this.
4x+3y=34
-1(4x+2y)=(28)-1
we can multiply the second equation by -1 to get y alone. (doesn't matter which equation). Once you do that, the positive 4x and the negative 4x cancel out, 3y-2y=1y and 34-28=6. you are left with
1y=6 so one shiny roll is 6 dollars.
now use that price to find the cost of the plain roll.
lets use Kathryn's equation
4x + 3(6)=34
4x + 18= 34
-18 -18
4x=16. Divide by 4 to find cost of one plain roll.
16÷4=4. One plain roll costs 4 dollars.
Lets check. Using Kathryn's equation,
4(4) + 3(6)=34
16+18=34
34=34. We are right.
Answer:
Step-by-step explanation:
The x value represents the time and y value represents the distance
