3 subtracted from 9 times 2 and 5 is added

Factor. 8x2y2 – 4x2y – 12xy 4(8x2y2 – x – 12xy) 4(2xy – 4x2y – 12xy) 4x2y2(2xy – xy –3) 4xy(2xy – x – 3)

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.

What other information do you need to prove triangle dac=bca by asa