Find the slope between the following points.
(2, 9) and (-3, 3)

m=-6

m=6/5

m=5/6

m=-6/5

Find the arc length parameter along the given curve from the point where tequals=0 by evaluating the integral s(t)equals=integral from 0 to t startabsolutevalue bold v left parenthesis tau right parenthesis endabsolutevalue d tau∫0tv(τ) dτ. then find the length of the indicated portion of the curve r(t)equals=1010cosine tcost iplus+1010sine tsint jplus+88t k, where 0less than or equals≤tless than or equals≤startfraction pi over 3 endfraction π 3.

Solve 3x-18=2y and 5x-6y=6 by elimination show work

Explain in a minimum of 2 sentences how to graph the equation of the absolute value function given a vertex of (-1,3) and a value of “a” equal to ½.