A (1.1), F 1 = (3.4), F 2 = (2. - 5). F 3 = (3.1), then the end point coordinate of the resultant force The options are (9,0) (0.9) (1.9) (9.1) In fact, I personally feel that the options are not right

A (1.1), F 1 = (3.4), F 2 = (2. - 5). F 3 = (3.1), then the end point coordinate of the resultant force The options are (9,0) (0.9) (1.9) (9.1) In fact, I personally feel that the options are not right

Vector AF1 = (2,3), vector af2 = (1, - 6), vector af3 = (2,0), so the resultant vector of vector AF1, af2 and af3 is (5, - 3), so the end coordinates of resultant force are (6, - 2)

If a / b = 3 / 5, then the value of (3a + 2b) / (a + 4b) is?

(a-2b)(-2b-a)-(3a+4b)(-3a+4b)

It is known that the quadratic function y = (T + 1) x squared + 2 (T + 2) x + 3 / 2 has the same function value when x = 0 and x = 2. If the image of the primary function y = KX + 6 is the same as that of the quadratic function

When x = 0 and x = 2 are equal, then 3 / 2 = 4 (T + 1) + 4 (T + 2) + 3 / 2
8t+12=0
t=-3/2
So the analytic formula is y = - x ^ 2 / 2 + X + 3 / 2
(2) A (- 3, m) is substituted to get m = - 9 / 2-3 + 3 / 2 = - 6
When a (- 3, - 6) is substituted into y = KX + 6, there is - 6 = - 3K + 6
That is, k = 4