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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.

Ade and abc are similar. which best explains why the slope of the line between points a and d is the same as the slope between points a and b?

Alina fully simplifies this polynomial and then writes it in standard form. xy2 – 2x2y + 3y3 – 6x2y + 4xy2 if alina wrote the last term as 3y3, which must be the first term of her polynomial in standard form? xy2 5xy2 –8x2y –2x2y