Asquare calendar has sides that are 9 inches long. what is the calendar area?

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.

Y=(x+3)(x-4)(x+1)^2 sulutions? multiplicity?

Find the missing dimension! need asap! tiny square is a =3025 in ²