What is the sum of (q+r)-(r+10)
1 answer:
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Answer:
f'(x) = 5[1 + f²(x)]
Step-by-step explanation:
By definition, f'(x) = lim h => 0 {[f(x + h) - f(x)]/h}
This was used, together with the given limits:
lim x => 0 f(x) = 0
lim x => 0 f(x)/x = 5,
to determine the derivative of the given function:
f(x + y) = [f(x) + f(y)]/[1 - f(x)f(y)].
The workings are shown in the attachments.
The areas of the parallelograms can be compared as: A. The area of parallelogram 1 is 4 square units greater than the area of parallelogram 2.
<h3>What is a parallelogram?</h3>A parallelogram refers to a geometrical shape and it can be defined as a type of quadrilateral and two-dimensional geometrical figure that has two (2) equal and parallel opposite sides.
<h3>How to calculate the area of a triangle?</h3>Mathematically, the area of a triangle can be calculated by using this formula:
Area = ½ × b × h
Where:
- b represents the base area.
- h represents the height.
Mathematically, the area of a rectangle can be calculated by using this formula;
A = LW
Where:
- A represents the area of a rectangle.
- l represents the length of a rectangle.
- w represents the width of a rectangle.
Next, we would determine the area of the two parallelograms as follows:
Area of parallelogram 1 = Area of red-rectangular figure - Area of triangle A - Area of triangle B - Area of triangle C - Area of triangle D.
Substituting the given parameters into the formula, we have;
Area of parallelogram 1 = (6 × 6) - (½ × 4 × 2) - (½ × 2 × 4)- (½ × 4 × 2) - (½ × 2 × 4)
Area of parallelogram 1 = 36 - 4 - 4 - 4 - 4
Area of parallelogram 1 = 36 - 16
Area of parallelogram 1 = 20 units².
For parallelogram 2, we have:
Area of parallelogram 2 = Area of blue-rectangular figure - Area of triangle P - Area of triangle Q - Area of triangle R - Area of triangle S.
Substituting the given parameters into the formula, we have;
Area of parallelogram 2 = (8 × 4) - (½ × 2 × 2) - (½ × 6 × 2)- (½ × 2 × 2) - (½ × 2 × 6)
Area of parallelogram 2 = 32 - 2 - 6 - 2 - 6
Area of parallelogram 2 = 32 - 16
Area of parallelogram 2 = 16 units².
Difference = Area of parallelogram 1 - Area of parallelogram 2
Difference = 20 - 16
Difference = 4 units².
In conclusion, we can infer and logically deduce that the area of parallelogram 1 is 4 square units greater than the area of parallelogram 2.
Read more on parallelogram here: brainly.com/question/4459854
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