The answer is 13, contains 1 tenth and a 3
-2 (x-2y=-5)
-2x+4y=10
2x+3y=4
0+7y=14
/7. /7
y=2
2x+3 (2)=4
2x+6=4
-6. -6
2x=-2
/2. /2
x=-1
If x represents the width of the poster (including borders), the area of the finished poster can be written as
.. a = x*(390/(x -10) +8)
.. = 8x +390 +3900/(x -10)
Then the derivative with respect to x is
.. da/dx = 8 -3900/(x -10)^2
This is zero at the minimum area, where
.. x = √(3900/8) +10 ≈ 32.08 . . . . cm
The height is then
.. 390/(x -10) +8 = 8 +2√78 ≈ 25.66 . . . . cm
The poster with the smallest area is 32.08 cm wide by 25.66 cm tall.
_____
In these "border" problems, the smallest area will have the same overall dimension ratio that the borders have. Here, the poster is 10/8 = 1.25 times as wide as it is high.
Answer:
Incorrect
Step-by-step explanation:
Juanita is incorrect because she shouldn't be using 2 squared.
Answer:
See Explanation
Step-by-step explanation:
Given
Required
Determine the number of cans for the wall
The dimension of the wall is not given. So, I will use the following assumed values:
First, calculate the area of the wall
If
Then 
Cross Multiply:
Make x the subject
400 cans using the assume dimensions.
So, all you need to to is, get the original values and follow the same steps
