The question is incomplete. Here is the complete question.
Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = -
x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.
Answer: L = 1; W = 9/4; A = 2.25;
Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:
A = x.y
A = x(-
)
A = -
To maximize, we have to differentiate the equation:
= 
(-)
= -3x + 3
The critical point is:
= 0
-3x + 3 = 0
x = 1
Substituing:
y = -x + 3
y = -.1 + 3
y = 9/4
So, the measurements are x = L = 1 and y = W = 9/4
The maximum area is:
A = 1 . 9/4
A = 9/4
A = 2.25
Answer:
l = 32.5 units, w= 27.5 units, A = 893.75 units²
Step-by-step explanation:
width is w
length is l = 5+w
P = 2( l+w) , substitute l for 5+w
P = 2(5+w+w)
P = 2(5+2w)
P = 10 +4w
P = 120
10 +4w = 120
4w = 120-10
4w = 110
w= 110/4
w= 27.5 units
l = 5+w = 5+ 27.5 = 32.5 units
A = l*w = 27.5 * 32.5 = 893.75 units ²
Answer:
You can break 24 things
Step-by-step explanation:
One minute is 60 seconds, so all you have to do is double how many things you can break in 30 seconds
Answer:
Standard Deviation = 5.928
Step-by-step explanation:
a) Data:
Days Hours spent (Mean - Hour)²
1 5 61.356
2 7 34.024
3 11 3.360
4 14 1.362
5 18 26.698
6 22 84.034
6 days 77 hours, 210.834
mean
77/6 = 12.833 and 210.83/6 = 35.139
Therefore, the square root of 35.139 = 5.928
b) The standard deviation of 5.928 shows how the hours students spend outside of class on class work varies from the mean of the total hours they spend outside of class on class work.
