Answer:
80
Step-by-step explanation:
every triangle, no matter what shape or size, will have all angles add up to 180 degrees so 180-40-60=80
2f(x) = 2x - 4 [0, 3]3f(x) = 3x - 1 [-2, -1]9f(x) = x² [4, 5]4f(x) = 4x [5, 20]5f(x) = x² - 3 [0, 5]1f(x) = x + 10 [-5, -1]10f(x) = 10x [-3, 0]¹/₂f(x) = 0.5x - 2 [2, 4]11f(x) = 2x² + x [1, 4]-1f(x) = -x + 2 [-3, 5]Domainthe set of all reasonable input values of x for the functionRangeset of output y values for the domain of the functionAverage Rate of ChangeChange in values over a given interval.Origin(0,0) on the coordinate graphing system; where the two axes meetx-axisthe horizontal number line in the coordinate systemy-axisthe vertical number line in the coordinate systemCoordinatesany specific (x,y) in the coordinate systemx-interceptwhere the function intersects the x-axisy-interceptwhere the function intersects the y-axis; the b value in a linear functionLinear FunctionA function whose graph is a straight line, where the average rate of change (slope) is constant.Exponential FunctionA function where the average rate of change is not constant and whose input value is an exponent.Table of ValuesA table showing two sets of related numbers<span>Slope of line through the points (-2, 3) and (0,0)
m = (0 - 3) / (0 - -2) = -3/2</span><span>Average Rate of Change on the interval
[-2, 0]</span>Slope: m = "rise over run" = 2Rate of Change<span>Slope of line through the points (5, -1) and (0,0)
m = (0 - -1) / (0 - 5) = -1/5</span><span>Average Rate of Change on the interval
[0, 5]</span><span>Slope of line through the points
(0, 16) and (4, 21)
m = (21 - 16) / (4 - 0) = 5/4</span>Average Rate of Change over the interval [0,4]
Answer:
8
Step-by-step explanation:
9.6 divided by 3/4 = 0.8, so 8. Correct me if I'm wrong ya'll.
Check the picture below.
let's recall that once we move an angle in the opposite direction of the 0-line or x-axis, we end up with a negative counterpart angle.
since we know that θ is on the IV Quadrant, then its negative counterpart must be across the line, and tangent is negative on the IV Quadrant, and positive on the I Quadrant.
