- Flight Of A Rocket Suppose The Altitude In Metres Of A Rocket T Seconds Into Flight Is Given By F T 2t 27t 1 (40.69 KiB) Viewed 67 times
- Flight Of A Rocket Suppose The Altitude In Metres Of A Rocket T Seconds Into Flight Is Given By F T 2t 27t 2 (25.84 KiB) Viewed 67 times
- Flight Of A Rocket Suppose The Altitude In Metres Of A Rocket T Seconds Into Flight Is Given By F T 2t 27t 3 (48.9 KiB) Viewed 67 times
- Flight Of A Rocket Suppose The Altitude In Metres Of A Rocket T Seconds Into Flight Is Given By F T 2t 27t 4 (72.87 KiB) Viewed 67 times
- Flight Of A Rocket Suppose The Altitude In Metres Of A Rocket T Seconds Into Flight Is Given By F T 2t 27t 5 (68.17 KiB) Viewed 67 times
Flight of a Rocket. Suppose the altitude (in metres) of a rocket t seconds into flight is given by f(t) = -2t³ + 27t² - 108t + 102 The interval where f(t) is concave up is O O (3, 6) 9 -∞, - (-∞0, 3) U (6, ∞) (-∞0, ∞0) 9 (²2/₁00) none of the other answers
Interest Formula. Suppose you work at a bank and the manager provides a new type of formula to calculate interest when merging two loans together. Suppose there are two loans with one interest rate at 5% and a second loan at 2%. The manager writes the first part of the formula for the new interest rate as (5in(2) + 2ln(5)) Which of the following is equivalent to the manager's expression? 0 0 0 0 0 0 00 400 e 400 0 e7 €32 +25 e 800 800 57
Flight of a Model Rocket (0, 18) (0, 37) (0, 56) When is the rocket rising? (Round your answers to the nearest integer.) (18, 37) (37, 56) (0, 18) h(t) = - (0, 37) (0, 56) When is it descending? (Round your answers to the nearest integer.) (18, 37) The height (in feet) attained by a rocket t sec into flight is given by the function 1 t³ + 18t² + 37t + 12 3 (37, 56) (t > 0).
Second Derivative Test. Consider the function f(x, y) = 2xy + 3ln(x) + 10y (a) The critical point of f(x, y) is O (-1,1) There is no critical point (1, e) (e, 1) O (-1/2,2) O (0,0) O (-5,3/10) (1, -1) (b) Use the second derivative test for functions of two variables to determine that in part (a): the critical point is not in the domain the second derivative test is inconclusive the critical point is a saddle point the critical point is a local maximum the critical point is a local minimum
Domain. The domain of g(x)= 13x² O 50 x + 9 can be written: O O O ²-52x +39 (-∞, -9) U (-9, -3) U (-3, ∞) (-1, -3) U (-3, 00) None of the other answers (-1, 3) U (3, ∞0) (-9, 1) U (1, 3) U (3, ∞0) (-∞, 3) U (3, 9) U (9, ∞) [-1, 9) U (9, ∞0) O (-∞0, -1) U (-1,00)