tessa can now use her graph in the example to find the coordinates of the vertices of figure G'H'I'J. How could she have founf these coordinates without drawing the reflection of figure GHIJ?

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.

What’s the domain of this function? fast

Solve the problem. a student earned grades of c, a, b, and a in four different courses. those courses had these corresponding numbers of credit hours: 4, 5, 1, and 5. the grading system assigns quality points to letter grades as follows: a = 4, b = 3, c = 2, d = 1, and f = 0. compute the grade point average (gpa) and round the result to two decimal places. 3.40 3.50 8.75 2.18