Question: *PLEASE GIVE JUST THE ANSWER - I.E: OPTION 1 / A,B,C,D* NO WORKING OR EXPLANATION* Question 18: fEvaluate / (-[x-2| +4) dx using the following

 *PLEASE GIVE JUST THE ANSWER - I.E: OPTION 1 / A,B,C,D*NO WORKING OR EXPLANATION*Question 18: \fEvaluate / (-[x-2| +4) dx using thefollowing graph. 4 3 -5 -4 -3 -2 -1 0 2 34 5 6 X -1 of " (-1x - 21 + 4) ox = 14 o ." ( - 1x - 21 +4 ) dx = 16 of" (-1x-21+4)dx = 10 O J .( -1x - 21 + 4 ) ox = 12Find the totalarea between the x-axis and the curve f(x) = - x-+4 betweenX = - 1 and x = 3. O 28 3 unitsO 29 3 - units? O 34 3 units? O 20 3units?Consider the graph below, which represents the velocity v(t) of a rowboat

*PLEASE GIVE JUST THE ANSWER - I.E: OPTION 1 / A,B,C,D* NO WORKING OR EXPLANATION*

Question 18:

moving away from its starting point onshore at 12 PM (t =0). A positive velocity indicates straight-line motion away from shore; while anegative velocity indicates straight- line motion back toward shore. Use the graphto find v(t) dt and determine the position of the boat relativeto 0 shore at 7 pm.\fO v(1)at = 12 miles; the boatis 12 miles away from shore at 7 PM. O v(t)dt =22 miles; the boat is 22 miles away from shore at 7PM. 0 O v(t)dt = - 10 miles; the boat is 10miles away from shore at 7 PM. 0 O v(t)dt = 10miles; the boat is 10 miles away from shore at 7 PM.025 ft 12 feet 25 ft 10 feet 15 feet = =-

\fEvaluate / (-[x-2| +4) dx using the following graph. 4 3 -5 -4 -3 -2 -1 0 2 3 4 5 6 X -1 of " (-1x - 21 + 4 ) ox = 14 o ." ( - 1x - 21 + 4 ) dx = 16 of" (-1x-21+4)dx = 10 O J . ( -1x - 21 + 4 ) ox = 12Find the total area between the x-axis and the curve f(x) = - x-+4 between X = - 1 and x = 3. O 28 3 units O 29 3 - units? O 34 3 units? O 20 3 units?Consider the graph below, which represents the velocity v(t) of a rowboat moving away from its starting point onshore at 12 PM (t = 0). A positive velocity indicates straight-line motion away from shore; while a negative velocity indicates straight- line motion back toward shore. Use the graph to find v(t) dt and determine the position of the boat relative to 0 shore at 7 pm.\fO v(1)at = 12 miles; the boat is 12 miles away from shore at 7 PM. O v(t)dt = 22 miles; the boat is 22 miles away from shore at 7 PM. 0 O v(t)dt = - 10 miles; the boat is 10 miles away from shore at 7 PM. 0 O v(t)dt = 10 miles; the boat is 10 miles away from shore at 7 PM. 025 ft 12 feet 25 ft 10 feet 15 feet = =- - - - 11 feet - - - = = =- 13 feet 8 feetUse the trapezoidal rule to approximate the surface area of the lake. O 1725 units O 1475 units O 3450 units O 1983.3 units

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