The cross-sectional view of a tunnel is shown on the axes below. The line [AB] representsa vertical wall located at the left side of the tunnel. The height, in metres, of the tunnel abovethe horizontal ground is modelled by y = -0.1 x 3 + 0.8 x 2 , 2 ≤ x ≤ 8 , relative to an origin O.yx2 4 6 8246A108EBCDOPoint A has coordinates (2 , 0), point B has coordinates (2 , 2.4), and point C hascoordinates (8 , 0).(a) (i) Find dy___dx .(ii) Hence find the maximum height of the tunnel. [6](b) Find the height of the tunnel when(i) x = 4 .(ii) x = 6 . [3](c) Use the trapezoidal rule, with three intervals, to estimate the cross-sectional area ofthe tunnel. [3](d) (i) Write down the integral which can be used to find the cross-sectional area ofthe tunnel.(ii) Hence find the cross-sectional area of the tunnel. [4]

Josh wants to know the length of a tunnel built through a mountain. To do so, he makes the measurements shown in the figure below. Use these measurements to find the length of the tunnel.

Define Tunnel and Elevated Railway.

Which ordered pair is not an endpoint on the shape shown?View Image DescriptionA coordinate plane with the points 𝐴=(−7,−2), 𝐵=(2,6), and 𝐶=(2,−5) plotted. There are line segments connecting the points A to B, B to C, C to A. These connections form a triangle.

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.

In which direction does the parabola x+8y2=2 open?