The area of a square is the square of the sides, and in this case, equals 14*16=224 mm^2
So length of each side
= sqrt(224) = 14. 97 or 15 mm to the nearest integer.
Answer:
For maximum area of the rectangular exercise run dimensions will be 50ft by 25ft.
Step-by-step explanation:
Let the length of the rectangular exercise run = l ft
and width of the run = w ft
Sinoman has to cover a rectangular exercise run from three sides with the fencing material,
So length of the material = (l + 2w) ft
l + 2w = 100
l = 100 - 2w --------(1)
Area of the rectangular area covered = Length × width
A = lw
A = w(100 - 2w) [(l = 100 - 2w)from equation (1)
For maximum area we find the derivative of area and equate it to zero.
![\frac{dA}{dw}=\frac{d}{dw}[w(100-2w)]](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdw%7D%3D%5Cfrac%7Bd%7D%7Bdw%7D%5Bw%28100-2w%29%5D)

A' = 100 - 4w
For A' = 0
100 - 4w = 0
4w = 100
w = 25 ft
From equation (1)
l = 100 - 2w
l = 100 - 2×(25)
l = 50 ft
Therefore, for maximum area of the rectangular exercise run dimensions will be 50ft by 25ft.
Answer:
D
Step-by-step explanation:
So here we have a line plot that shows lengths of insects and the quantity of insects that fits the description.
For instance, there are 4 Xs for 1/8. This means that there are four insects that are 1/8 inches long. Likewise, there is only 1 X for 7/8. This means that are is only one insect that is 7/8 inches long.
To find out how much longer is the longest insect than the shortest insect, we simply need to subtract the shortest into the longest.
The longest insect we have is the lone insect who is 7/8 inches long.
And the shortest insects we have are the four insects who each measure 1/8. However, we only need 1.
So, subtract:

Combine fractions with common denominators:

Subtract:

The answer is D.
All you would do is divide for this problem so,
13.75 ÷ 5 = 2.75
So one pack would cost 2.75
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