What is wrong with the equation? 3 x−5 dx = x−4 −4 3 −1 = 20 81 −1 There is nothing wrong with the equation. f(x) = x−5 is continuous on the interval [−1, 3] so FTC2 cannot be applied. The lower limit is less than 0, so FTC2 cannot be applied. f(x) = x−5 is not continuous on the interval [−1, 3] so FTC2 cannot be applied. f(x) = x−5 is not continuous at x = −1, so FTC2 cannot be applied.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

step-by-step explanation:

points a and b are equidistant from the desired line, so the line you find by using them will go through point r.

using points other than a and b will give the desired line only if they are equidistant from r. if they are randomly located, the perpendicular line you create will not necessarily go through point r, hence will not be a solution to the problem.

1. 2

2. 11

3. 5

4. 6

5. 7

6. 5

7. -1

8. 1

step-by-step explanation: