Chapter 1.1, Problem 56E

temperature in degrees Fahrenheit, n hours since midnight. 5. The temperature was recorded at several times during the day. Function T gives the Here is a graph for this function. To 29uis a. Describe the overall trend of temperature throughout the day. temperature (Fahrenheit) 40 50 50 60 60 70 5 10 15 20 25 time of day b. Based on the graph, did the temperature change more quickly between 10:00 a.m. and noon, or between 8:00 p.m. and 10:00 p.m.? Explain how you know. (From Unit 4, Lesson 7.) 6. Explain why this graph does not represent a function. (From Unit 4, Lesson 8.)

Find the area of the shaded region. (a) 5- y 3 2- (1,4) (5,0) 1 3 4 5 6 (b) 3 y 2 Decide whether the problem can be solved using precalculus, or whether calculus is required. If the problem can be solved using precalculus, solve it. If the problem seems to require calculus, use a graphical or numerical approach to estimate the solution. STEP 1: Consider the figure in part (a). Since this region is simply a triangle, you may use precalculus methods to solve this part of the problem. First determine the height of the triangle and the length of the triangle's base. height 4 units units base 5 STEP 2: Compute the area of the triangle by employing a formula from precalculus, thus finding the area of the shaded region in part (a). 10 square units STEP 3: Consider the figure in part (b). Since this region is defined by a complicated curve, the problem seems to require calculus. Find an approximation of the shaded region by using a graphical approach. (Hint: Treat the shaded regi as…

Solve this differential equation: dy 0.05y(900 - y) dt y(0) = 2 y(t) =

Complete the description of the piecewise function graphed below. 6 5 -7-6-5-4-3-2-1 2 3 5 6 -1 -2 -3 -4 -5 { f(x) = { { -6 if -6x-2 if -2< x <1 if 1 < x <6

Let F = V where (x, y, z) x2 1 + sin² 2 +z2 and let A be the line integral of F along the curve x = tcost, y = t sint, z=t, starting on the plane z = 6.14 and ending on the plane z = 4.30. Then sin(3A) is -0.598 -0.649 0.767 0.278 0.502 0.010 -0.548 0.960

Let C be the intersection of the cylinder x² + y² = 2.95 with the plane z = 1.13x, with the clockwise orientation, as viewed from above. Then the value of cos (₤23 COS 2 y dx xdy+3 z dzis 3 z dz) is 0.131 -0.108 -0.891 -0.663 -0.428 0.561 -0.332 -0.387