section 1_10 solutions
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section 1_10 solutions

Course Number: MATH math 150, Spring 2009

College/University: Ohio State

Word Count: 849

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Math 1650 Homework Solutions Jason Snyder, PhD. 1.10 2-36 (even), 42 52 (even), 54, 56, 60, 68 Solutions 1 8 Find the slope of the line through P and Q. 2) P(0,0), Q(2,-6) = 2 - 1 2 - 1 = -6-0 2-0 = -6 2 = -3. 4) P(1,2), Q(3,3) = 2 - 1 2 - 1 = 3-2 3-1 = . 2 1 6) P(2,-5), Q(-4,3) = 2 - 1 2 - 1 = 3-(-5) -4-2 = 3+5 -6 = 8 -6 =- . 3 4 8) P(-1,-4), Q(6,0) = 2 - 1 2 - 1 = 0-(-4)...

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1650 Math Homework Solutions Jason Snyder, PhD. 1.10 2-36 (even), 42 52 (even), 54, 56, 60, 68 Solutions 1 8 Find the slope of the line through P and Q. 2) P(0,0), Q(2,-6) = 2 - 1 2 - 1 = -6-0 2-0 = -6 2 = -3. 4) P(1,2), Q(3,3) = 2 - 1 2 - 1 = 3-2 3-1 = . 2 1 6) P(2,-5), Q(-4,3) = 2 - 1 2 - 1 = 3-(-5) -4-2 = 3+5 -6 = 8 -6 =- . 3 4 8) P(-1,-4), Q(6,0) = 2 - 1 2 - 1 = 0-(-4) 6-(-1) = . 7 1 2 4 10) (a) Sketch lines through (0,0) with slopes 1, 0, , 2, and - 1. (b) Sketch lines through (0,0) with slopes , , - , and 3. 3 2 3 1 1 1 y y 5 5 x -8 -6 -4 -2 2 4 6 8 x -8 -6 -4 -2 2 4 6 8 -5 -5 Page 1 of 9 Math 1650 Homework Solutions Jason Snyder, PhD. 11 14 Find an equation for line whose graph is sketched. 12) The line passes through the points (-2,0) and (0,4), thus its slope is given by = 4-0 4 = = 2. 0 - (-2) 2 The slope-intercept form of the line is therefore given by = 2 + 4. 14) The line passes through the points (-3,0) and (0,-4), thus its slope is given by = -4 - 0 4 =- . 0 - (-3) 3 The slope-intercept form of the line is therefore given by 4 = - - 4. 3 Page 2 of 9 Math 1650 Homework Solutions Jason Snyder, PhD. 15 34 Find an equation of the line that satisfies the given conditions. 16) Through (-2,4); slope -1 - 4 = -1( + 2) 18) Through (-3,-5); slope - 7 2 7 + 5 = - ( + 3) 2 20) Through (-1,-2) and (4,3) = 3 - (-2) 5 = =1 4 - (-1) 5 + 2 = + 1 22) slope ; y-intercept 4 5 2 = 24) x-intercept -8; y-intercept 6 2 +4 5 This line passes through the points (-8,0) and (0,6) = 6-0 6 3 = = 0 - (-8) 8 4 3 = + 6 4 Page 3 of 9 Math 1650 Homework Solutions Jason Snyder, PhD. 26) Through (4,5); parallel to x-axis. Parallel to x-axis means m = 0, so the line is horizontal. = 5 28) y-intercept 6; parallel to the line 2x+3y+4=0 The line 2x + 3y + 4 = 0 written in slope-intercept form becomes 2 4 = - - 3 3 So the slope of any parallel line will be = - . 3 2 2 = - + 6 3 30) Through (2,6); perpendicular to the line y = 1 The line y = 1 is horizontal, so any line perpendicular to it must be vertical. = 2 32) Through (1,7); perpendicular to the line 4x 8y = 1 The line 4x 8y = 1 written in slope-intercept form becomes 1 1 = - - 2 8 So the slope of any perpendicular line will be m = 2. - 7 = 2 - 1 Page 4 of 9 Math 1650 Homework Solutions Jason Snyder, PhD. 34) Through (-2,-11); perpendicular to the line passing through (1,1) and (5,-1) The line passing through (1,1) and (5,-1) has slope = -1 - 1 2 1 =- =- . 5-1 4 2 So the slope of any perpendicular line will be m = 2. + 11 = 2( + 2) 36) (a) Sketch the line with slope -2 that passes through the point (4,-1). (b) Find an equation for this line. y 8 6 4 2 x -8 -6 -4 -2 -2 2 4 6 8 -4 -6 -8 + 1 = -2 - 4 41 52 Find slope the and y-intercept of the line and draw its graph 42) 3x 2y = 12 Re-writing in slope-intercept form, we get = - 6 2 3 Slope ; y-intercept -6 2 3 Page 5 of 9 Math 1650 Homework Solutions Jason Snyder, PhD. 44) 2x 5y = 0 Re-writing in slope-intercept form, we get 2 = 5 Slope ; y-intercept 0 5 2 46) -3x -5y +30 = 0 Re-writing in slope-intercept form, we get 3 = - + 6 5 Slope - ; y-intercept 6 5 3 48) 4y + 8 = 0 Re-writing in slope-intercept form, we get = -2 Slope 0; y-intercept -2 50) x = -5 Undefined slope; no y-intercept Page 6 of 9 Math 1650 Homework Solutions Jason Snyder, PhD. 52) 4x + 5y = 10 Re-writing in slope intercept form, we get = - + 2 5 4 Slope - ; y-intercept 2 5 4 54) Use slopes to show that A(-3,-1), B(3,3), and C(-9,8) are vertices of a right triangle. The slope of the line passing through A and B is = 3 - -1 4 2 = = 3 - -3 6 3 The slope of the line passing through B and C is = 8-3 5 =- -9 - 3 12 The slope of the line passing through C and A is = 8 - (-1) 9 3 =- =- -9 - (-3) 6 2 Since = -1, we conclude that that the lines AB and CA are perpendicular. Thus the points A, B, and C are the vertices of a right triangle. Page 7 of 9 Math 1650 Homework Solutions Jason Snyder, PhD. 60) (a) Find an equation for the line tangent to the circle x2 + y2 = 25 at the point (3,-4). (b) At what other point on the circle will the tangent line be parallel to the tangent line in part (a)? The line passing through the origin and the point (3,-4) has slope = - 4 3 3 4 Any line perpendicular to this line will have slope = . (a) + 4 = ( - 3) 4 3 (b) (-3,4) Page 8 of 9 Math 1650 Homework Solutions Jason Snyder, PhD. 68) Depreciation A small business buys a computer for $4000. After 4 years the value of the computer is expected to be $200. For accounting purposes, the business uses linear depreciation to assess the value of the computer at a given time. This means that if V is the value of the computer at time t, then a linear equation is used to relate V and t. (a) Find a linear equation that relates V and t. (b) Sketch a graph of this linear equation. (c) What do the slope and V-intercept of the graph represent? (d) Find the depreciated value of the computer 3 years from the date of purchase. (a) We know that the line representing the depreciated value of the computer will pass through the points (0,4000) and (4,200). Thus the slope of this line is = 200 - 4000 = -950. 4 So the equation of the line, in slope-intercept form is = -950 + 4000. (b) Sketch of graph not given here (c) the slope m = -950 means that the computer's value is decreasing at a rate of $950 per year. The V-intercept represents purchase price of the computer. (d) When t = 3, we have = -950 3 + 4000 = -2850 + 4000 = $1150 Page 9 of 9

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