Question

POINTS POINTS POINTS!!! PLS HELP QUICKLY!!! SIMPLE MATH!!

The function g is defined by g(x)=9k−4, where k is a constant. Find k, if the graph of g passes through the point (7,−2).

Answer 1

Using the parameter s = t3 we get:

t

3, −7t

3, t3

= s, −7s,s = s 1, −7, 1

This is a parametrization of the line through the origin, with the direction vector v = −1, 7, 1.

(c) The parametrization

8 − 4t3, 2 + 5t2, 9t3

does not parametrize a line. In particular, the points (8, 2, 0) (at t = 0),

(4, 7, 9) (at t = 1), and (−24, 22, 72) (at t = 2) are not collinear.

2. What is the projection of r(t) = ti + t4j + etk onto the xz-plane?

solution The projection of the path onto the xz-plane is the curve traced by ti + etk =

t, 0, et

. This is the curve

z = ex in the xz-plane.

3. Which projection of cost, cos 2t,sin t is a circle?

solution The parametric equations are

x = cost, y = cos 2

Answer 2

Graph of g(x)=9k−4 passes thru (7, -2). Find k.

-2 =9k - 4, or 2 = 9k, or k = 2/9

However, I notice that your g(x)=9k−4 does not have "x" in the right side. That would mean that g(x) is a constant equal to -2,

Please go back and check whether or not your g(x)=9k−4 was copied down correctly. Is it possible that you meant g(x)=9kx−4??