The vertices of a quadrilateral in the coordinate plane are known. How can the perimeter of the figure be found? Use the distanc
2 answers:
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Answer:
positive integers = x
4.9 < x < 10.1
x = 5, 6, 7, 8, 9, 10
sum = 5 + 6 + 7 + 8 + 9 + 10
= 45
If the rectangle has a fixed width of 300, it means that
where and
are width and length, respetively.
So given the length , the are would be
A) The signs of the first derivative (g') tell you the graph increases as you go left from x=4 and as you go right from x=-2. Since g(4) < g(-2), one absolute extreme is (4, g(4)) = (4, 1).
The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).
b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.
c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.
The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).
b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.
c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.