Accuplacer Math Terms Exam Bank Solution Manual Already Passed (Graded A+)

Accuplacer Math Terms Exam Bank Solution Manual Already Passed (Graded A+) Define - "ordered pairs" - Answers = (x, y) = a single point on a coordinate plane Define - "origin" - Answers = (0,0) = the point where two axes meet Explain - "quadrants" - Answers = when the two axes (x and y) cross they form 4 quadrants Explain what - "the distance" means Define - "distance formula" for a line(s) - Answers = "distance" - the length between two points on a plane - (x1, y1), (x2, y2) * does not have to be forming a line *distance is ALWAYS positive = (Distance (d) = "square root" all over - [( x2-x1)^2 + (y2-y1)^2] Explain what - "the midpoint" means Define - "midpoint formula" for a line(s) - Answers = "midpoint" - the middle point between two points on a plane - (x1, y1), (x2, y2) -that are forming one line *remember - "MidPOINT" = Midpoint (M) = [(x1+x2)/2] , [(y1+y2)/2] Coordinate Plane = ______________ demential. - Answers = 2D Describe/Define what a "one demential (1D) line" is In a 1D line, what two things need to be accounted for and explain how we can do this - Answers = a horizontal or vertical line = (x=0) - horizontal - (look at the location of (x) on the coordinate plane) = (y=0) - vertical - (look at the location of (y) on the coordinate plane) = Distance and Midpoint: = horizontal = (D = ("absolute value" X2-x1)) = (M = (x1+x2)/2) = vertical = (D = ("absolute value" y2-y1)) = (M = (y1+y2)/2) Define/Explain - "slope" of a line - Answers = (m)=(rise/run)=(change in y/change in x) = (y2-y1/x2-1) When looking at a graphed line - how can we immediately know if the slope is negative or positive - Answers = (-) "negative slope" = (line starts in Q3 - the line ends in Q1) = (+) "positive slope" = (lines starts in Q1 - the line ends in Q4) Horizontal Lines have ____________ slope. - Answers = (0) Vertical Lines have ____________ slope. - Answers = "undefined", "no slope", "infinite" Give the - Standard Form of a Linear Equation Describe how to quickly find the - "slope" and "y - intercept" of this specific equation - Answers = (Ax + By = C) *rule: A or B can NOT = (0) *rule: A can NOT be "negative" (-) = "slope" = (-A/B) = "y-intercept" = (C/B) Give the - Standard Form of a Quadratic Equation - Answers = (ax^2 + bx + c = 0) Give the - General Form of a line - and describe what it essential means - Answers = (Ax + By +C = 0) *rule: A can NOT = (0) = "some expression" [= (0)] - expression ALWAYS equals 0 Describe what the "Standard Form of a Polynomial means" - Answers = equation is ordered from (largest - smallest) according to the coefficients degrees Define what the "x-intercept" is - Answers = point of the (x) axis = (y=0) Define what the "y-intercept" is - Answers = point of the (y) axis = (x=0) Define the - Point-Slope Form of a line What is usually given in the problem that triggers to use this specific form - Answers = given = slope (m) and a point (x1, y1) *remember: the form of the line is based off what is given - "point-slope" = [m = (y - y1) / (x - x1)] Define the - Slope-Intercept Form of a line What is usually given in the problem that triggers to use this specific form - Answers = given = slope (m) and the y-intercept (b) *remember: the form of the line is based off what is given - "slope-intercept" = (y=mx+b) What kind of a line is defined as - (y=mx) - Answers = a line that passes through the origin Define - the "slope" of 2 - Parallel Line & - the "slope" of 2 - Perpendicular Line - Answers = 2 parallel line's slopes = SAME = 2 perpendictular line's slopes = NEGATIVE RECIPROCAL : [(1/x) - (x/-1)] How would you - "solve by graphing" these two lines: (2x+y=8) and (2x+3y=12) - Answers = graph by using - "intercepts" = find the (x) & (y) intercept for the first line: - set (x) in the first equation to (x=0) - then solve for (y), so now you have (0,y) - which is the "y-intercept" - set (y) in the first equation to (y=0) - then solve for (x), so now you have (x,0) - which is the "x-intercept" - take the points (x,0) and (0,y) and graph the first line - repeat the steps above for the second line equation = now that you have two lines graphed you can find the - "point at which they meet " = [(P) = (x,y)] - ANSWER - you can check your answer by plugging it into the two given equations Define - (h,k) - Answers = (h,k) = the "center" point of a circle - a "fixed point" Define what the - distance from (h,k) to any (x,y) point is - Answers = (r) = radius State the - "Radius (r) Formula" - Answers = Radius (r) = "square root all over" : [(x-h)^2 (y-k)^2] According to the - "Equation of a Circle" - define (x,y) - Answers = (x, y) = any point on the circumference of a circle Give the - "Equation of a Circle" - Answers = [(r^2) = (x-h)^2 + (y-k)^2] Define - "Straight Angle" - Answers = 180 (degrees) - (singular) Define - "Reflex Angle" - Answers = MORE - than 180 (degrees) - but LESS - than 360 (degrees) - (singular) Define - "Supplementary Angles" - Answers = angles that "add" up to = 180 (degrees) - (plural) Define - "Complementary Angles" - Answers = angles that "add" up to = 90 (degrees) - (plural) Define what the term - "congruent angles" means - Answers = angles that are the SAME in -