(Easy) An object is launched at 19.6 meters per second (m/s) from a 5.58 meter tall platform. The equation for the object’s height s at time t seconds after launch is s(t) = 4.9t^2 + 19.6t + 58.8s(t) = -4.9^2 + 19.6t + 58.8, where s is in meters. How high will the object be after 5 seconds?

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.

Use δjkl and δmpn to determine whether don's claim is true or false. which of the following should don use to prove the triangles are similar? sss sas aas aa

Ialready asked this but i never got an answer. will give a high rating and perhaps brainliest. choose the linear inequality that describes the graph. the gray area represents the shaded region. y ≤ –4x – 2 y > –4x – 2 y ≥ –4x – 2 y < 4x – 2