The slope of the given line is y2−y1x2−x1 = 6−45−3 = 1
Therefore 4−(−1)3−x = 1
That is 4+1 = 3 - x
Therefore x = -2
∼(p→(q∧r))
=p∧∼(q∧r) since ∼(p→q)≡p∧∼q
=p∧(∼q∨∼r)
sin(cot−1x) is equal to
Let f (x + y) = f(x) + f(y) ∀ x, y ∈R Suppose that f (6) = 5 and f ‘ (0) = 1, then f ‘ (6) is equal to