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: About 12.083
Explanation:
In order to solve this you would use Pythagorean theorem.
You are given the length of the two legs (5 and 11)
So by plugging in 5 and 11 into A Squared plus B Squared = C Squared, you get that C^2= 146.
By finding the square root of both sides you get square root of 146, which is approximately 12.803
Answer:
12 girls.
Step-by-step explanation:
24 ÷ 2 = 12
(This is divided to find out 1 part of the whole class as ⅓)
⅓ = 12
1 (whole) take away ⅔ = ⅓
Since ⅔ + ⅓ = 1
⅓ = fraction of girls in class
1 part = 12
So ⅓ = 12 girls out of 36 total students
Hope this helps. :)