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 one is triangle and the second one is rectangle
Step-by-step explanation:
i just did it on edge
Solve the area of the whole circle since in an hour, the minute hand covers the whole clock
assuming that the minute hand reaches all the way out to the edge of the clock
the minute hand is the radius
area=pi times r^2
r=5
5^2=25
area=pi times 25
area=25pi
aprox pi=3.14
25 times 2.14=78.5
area=78.5 in^2
Answer:
T4 = -2°F
Step-by-step explanation:
Given the following data;
Current temperature = 10°F
Constant, K = 3°F
Time, t = 4 hours
In four (4) hours, we have;
T = kt
Substituting into the equation, we have;
T = 3 * 4
T = 12°F
T4 = current temp - new temp
T4 = 10°F - 12°F
T4 = -2°F
Therefore, the temperature it will be in 4 hours is - 2°F.
Answer:
x = number of nickels = 127
y = number of dimes = 156
z = number of quarters = 78
Step-by-step explanation:
Let
x = number of nickels
y = number of dimes
z = number of quarters
Total worth of the coins = $41.45
Total number of coins = 361
x + y + z = 361 (1)
dime = $0.1,
nickel = $0.05
quarter = $0.25
0.05x + 0.1y + 0.25z = 41.45 (2)
twice as many dimes as quarters.
y = 2z
Substitute y = 2z into (1) and (2)
x + 2z + z = 361
0.05x + 0.1(2z) + 0.25z = 41.45
x + 3z = 361
0.05x + 0.2z + 0.25z = 41.45
x + 3z = 361 (3)
0.05x + 0.45z = 41.45 (4)
Multiply (4) by 20
x + 3z = 361 (3)
x + 9z = 829 (5)
Subtract (3) from (5)
9z - 3z = 829 - 361
6z = 468
Divide both sides by 6
z = 468 / 6
= 78
z= 78
Recall,
y = 2z
= 2(78)
= 156
y = 156
Substitute the value of y and z into
x + y + z = 361
x + 156 + 78 = 361
x + 234 = 361
x = 361 - 234
= 127
x= 127
x = number of nickels = 127
y = number of dimes = 156
z = number of quarters = 78