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 correct option is 4.
Step-by-step explanation:
The non parallel sides of an isosceles trapezoid are congruent.
The image of an isosceles trapezoid is same as the preimage of isosceles trapezoid if
1. Reflection across a line joining the midpoints of parallel sides.
2. Rotation by 360° about its center.
3. Rotation by 360° about origin.
If we rotate the trapezoid by 180° about its center, then the parallel sides will interchanged.
If we reflect the trapezoid across a diagonal, then the resultant figure will be a parallelogram.
If we reflect across a line joining the midpoints of the nonparallel sides, then the parallel sides will interchanged.
After rotation by 360° about the center, we always get an onto figure.
Therefore option 4 is correct.
Answer:
36- degree rotation m8
Step-by-step explanation:
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