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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.

Which shows the correct substitution of the values a,b and c from the equation 1=-2x+3x^2+1

Consider this equation. |y + 6| = 2 what can be concluded of the equation? check all that apply. there will be one solution. there will be two solutions. the solution to –(y + 6) = 2 will be also be a solution to the given absolute value equation. the solution(s) will be the number(s) on the number line 2 units away from –6. the value of y must be positive since the variable is inside absolute value signs.