Someone please help me with this

I've been working on this for a few days and i just don't understand, it's due in a few hours. you. the direction of a vector is defined as the angle of the vector in relation to a horizontal line. as a standard, this angle is measured counterclockwise from the positive x-axis. the direction or angle of v in the diagram is α. part a: how can you use trigonometric ratios to calculate the direction α of a general vector v = < x, y> similar to the diagram? part b suppose that vector v lies in quadrant ii, quadrant iii, or quadrant iv. how can you use trigonometric ratios to calculate the direction (i.e., angle) of the vector in each of these quadrants with respect to the positive x-axis? the angle between the vector and the positive x-axis will be greater than 90 degrees in each case. part c now try a numerical problem. what is the direction of the vector w = < -1, 6 > ?

Drag the tiles to the correct boxes to complete the pairs. match the exponential functions with their horizontal asymptotes and y-intercepts. f(x) = 7x − 4 f(x) = 3x+2 + 4 f(x) = 9x+1 − 4 f(x) = 2x + 4 horizontal asymptote and y-intercept exponential function horizontal asymptote: y = 4 y-intercept: (0, 5) arrowboth horizontal asymptote: y = -4 y-intercept: (0, 5) arrowboth horizontal asymptote: y = -4 y-intercept: (0, -3) arrowboth horizontal asymptote: y = 4 y-intercept: (0, 13) arrowboth

Find the value of x to the nearest tenth.