Solve for both x and y in the parallelogram below.

Find the value of 1715⋅211.though these numbers aren't quite as nice as the ones from the example or the previous part, the procedure is the same, so this is really no more difficult. give the numerator followed by the denominator, separated by a comma.

Find parametric equations for the line tangent to the curve of intersection of the surfaces at the given point. surfaces: xsquared2plus+2yplus+2zequals=1212 yequals=33 point: left parenthesis 1 comma 3 comma five halves right parenthesis

If sec theta = 5/3 and the terminal point determined by theta is in quadrant 4, then