Experison form I hope this helps
<h2>
Maximum area is 25 m²</h2>
Explanation:
Let L be the length and W be the width.
Aidan has 20 ft of fence with which to build a rectangular dog run.
Fencing = 2L + 2W = 20 ft
L + W = 10
W = 10 - L
We need to find what is the largest area that can be enclosed.
Area = Length x Width
A = LW
A = L x (10-L) = 10 L - L²
For maximum area differential is zero
So we have
dA = 0
10 - 2 L = 0
L = 5 m
W = 10 - 5 = 5 m
Area = 5 x 5 = 25 m²
Maximum area is 25 m²
Step-by-step explanation:
5x-7+3x+27=180
8x+20=180
8x=160
x=20°.
hope this helps you.
Digit 4 in 49,308 is ten thousands and the 4in 4061 is thousands
Hii!
So I did this and I have A!
A: You cannot get the mean from the graph but you CAN get the third quartile!
B: To find the interquartile range (IQR) we subtract the third and first quartiles:
60-35 = 25
C: An outlier would be much larger than the rest of the data or much smaller than the rest of the data. An outlier would make the "whisker" portion longer and could potentially slightly shift the box.
