3(x+5)2-100=92

The two after the )5 is an exponent. Help please!

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.

Jim had 3,067.48 in his checking account he wrote a check to pay for two airplane tickets. his account now has 1,845.24.

The graph of a curve that passes the vertical line test is a: