Number two is quadrant 3 if you look up quadrants on a graph it will show you where each quadrant is. You can use this website symbolab for number 3.
You can use Desmosgraphing for number 4
the answer is rate hope it helps
Answer:
part A: expression 1: 6(8m+2)
expression 2: 6m+12+42m
Part B: 6(m+2+7m)= 6(8m+2)
combine like terms is 6(m+2+7m), so it is 6(2+8m) or 6(8m+2)
6(8m+2)= 6(8m+2)
Part C: 6m+12+42m=6(m+2+7m)
m=0
6(0+2+7(0))=6(0)+12+42(0)
6(2)= 12
12=12
Yes, because it is continuous on [0,2] and differentiable on (0,2), the theorem states that there must exist some value c where a line tangent to c is parallel to the secant line through 0 and 2.
To solve for S, you would want to get rid of the denominator so multiply 360 onto both sides. then the equation is 360A=pi times r squared times S. Divide both sides by pi and r squared and you get S=360A/pi times r squared
