M23 113
6.
7.
144
mZ1 =
mZ2 =
m23 =
147
5
52
m24
m25 =
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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.

Ivan began dividing g2  – 6 by g + 1, noting that . he started by placing g in the quotient, as shown below.what is g2  – 6 divided by g + 1? g –  1 –  g + 1  – g  –  7 + g  –  5  + 

Line gh passes through points (2, 5) and (6, 9). which equation represents line gh? y = x + 3 y = x – 3 y = 3x + 3 y = 3x – 3