A curve in the xy-plane is defined by the vector valued function f, of, tf(t) equals= open angle bracket, x, left bracket, t, right bracket, comma, y, left bracket, t, right bracket, close angle bracket⟨x(t),y(t)⟩, where x, left bracket, t, right bracketx(t) equals= 5, cosine, left bracket, minus, 4, t, right bracket5cos(−4t) and y, left bracket, t, right brackety(t) equals= t, to the power minus 4 , plus, 3t −4 +3 for t, is greater than, 0t>0. What is start fraction, d, squared, y, divided by, d, x, squared, end fraction dx 2 d 2 y​ in terms of tt?

The graph of a function g is shown.The x y-coordinate plane is given. A function labeled y = g(x) with 4 parts is graphed.The first part is a curve, enters the window in the second quadrant, goes up and right becoming less steep, crosses the y-axis at approximately y = 2.5, and ends at the open point (2, 3).The second part is a curve begins again at the open point (2, 1), goes up and right becoming less steep, and ends at the open point (5, 2).The third part is the closed approximate point (5, 1.2).The fourth part is a curve, begins at the open point (5, 2) goes down and right becoming more steep, and exits the window in the first quadrant.Use it to state the values (if they exist) of the following:(a)  lim x → 2− g(x)(b)  lim x → 2+ g(x)(c)  lim x → 2 g(x)(d)  lim x → 5− g(x)(e)  lim x → 5+ g(x)(f)  lim x → 5 g(x)SolutionLooking at the graph we see that the values of g(x) approach as x approaches 2 from the left, but they approach as x approaches 2 from the right.Therefore (a) lim x → 2− g(x) =     and    (b) lim x → 2+ g(x) = .Since the left and right limits are different, we conclude that (c) the limit as x approaches 2 of g(x) does not exist.The graph also shows that (d) lim x → 5− g(x) =     and    (e) lim x → 5+ g(x) = .This time, the left and right limits are the same and so, by this theorem, we have (f) lim x → 5 g(x) = Despite this fact, notice that g(5) ≠ 2.

The Lorenz curve is used to represents

The x y-coordinate plane is given. A function composed of several line segments is on the graph. The function begins at y = −1 on the negative y-axis, goes up and right in a linear fashion, crosses the x-axis at x = 2, sharply changes direction at (4, 1), goes down and right in a linear fashion, sharply changes direction at (6, 0), goes up and right in a linear fashion, sharply changes direction at (8, 2), goes horizontally right, sharply changes direction at (12, 2), goes down and right in a linear fashion, sharply changes direction at (14, 1), goes up and right in a linear fashion, and ends at (16, 3).

The graph of a function g is shown. The x y-coordinate plane is given. The curve begins at the point (−2, 0), goes up and right, passes through the point (−1.5, 1), goes up and right, changes direction at the point (−1, 1.5), goes down and right, passes through the point (−0.5, 1), goes down and right, passes through the origin, goes down and right, passes through the point (0.5, −1), goes down and right, changes direction at the point (1, −1.5), goes up and right, passes through the point (1.5, −0.5), goes up and right, changes direction at the point (2, 0.5), goes down and right, crosses the x-axis at x = 2.5, goes down and right, changes direction at the point (3, −1), goes up and right, passes through the point (3.5, −0.5), goes up and right, and ends at the point (4, 0.5). Estimate 4 −2 g(x) dx with six subintervals using the following. (a) right endpoints 0 Correct: Your answer is correct. (b) left endpoints -0.5 Correct: Your answer is correct. (c) midpoints

A curve is symmetric about the  if the equation  r=f(θ) is unchanged when replacing  r with  −r, or  θ with  π+θ.r=f(θ). Similarly for every point  (r,θ) on the graph, the point  is also on the graph.