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:
The first option.
Step-by-step explanation:
In order to effectively simplify, you need to divide the side with the variable by a number that is less than or equal to the number. 3 goes into 36 twelve times.
3g = 36.
3g/3 = g
36/3 = 12
g = 12 when divided by 3.
First you need to set up an equation y = mx + b. m would be the monthly charge and b would be the one time fee. x represents the number of months.
To solve for the number of months with a total price of 240 we substitute 240 in for y and solve for x.
240 = 25x + 40
subtract 40 from each side
200= 25x
divide by 25
8 = x
8 months
10 + 7r = 45
7r = 35
r = 5
5 miles
What is the rest of the expression?
Answer:
70 + 4D
Step-by-step explanation:
- (2 x 100) + (7 x 10) + (4 x D) + (2 x 100)
- (200) + (70) + 4D + 200
-200 + 70 +4D + 200
-130 + 4D + 200
70 + 4D
