Find the gradient of the line joining (3,7) and (6,9). hence, find the acute angle it makes with the positive x-y axis

33.7 degrees

Step-by-step explanation:

As we go from (3,7) to (6,9), x increases by 3 and y increases by 2.  Thus, the gradient (slope) of the line connecting these two points is

m = rise / run = 2/3.  Using the slope-intercept formula y = mx + b, we obtain

7 = (2/3)(3) + b, or 7 = 2 + b, so we see that b = 5 and y = (2/3)x + 5.  The y-intercept is (0, 5).

Next we find the x-intercept.  We set y = (2/3)x + 5 = to 0 and solve for x:

(2/3)x = -5, or (3/2)(2/3)x = -5(3/2), or x = -15/2, so that the x-intercept is

(-15/2, 0).  This line intersects the x-axis at (-15/2, 0).

Now look at the segment of this line connecting (-15/2, 0) and (0, 5).  Here x increases by 15/2 and y increases by 5, and so the tangent of the acute angle in question is

tan Ф = 5 / (15/2) = 10 / 15 = 2/3.

Using the inverse tangent function, we get Ф = arctan 2/3, or approx.

33.7 degrees.

I believe you meant "the acute angle it makes with the positive x-axis."

i think is g(1)=-1.

hope that i can