course Mth 158
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RESPONSE --> x - 2 y = -5 (0,0) -2y = -x - 5 y = 1/2x + 5/2 Slope for parallel line is 1/2 y = 1/2x
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17:26:22 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line parallel to this will therefore have slope 1/2. Point-slope form gives us y - 0 = 1/2 * (x - 0) or just y = 1/2 x. **
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RESPONSE --> ok
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17:27:36 **** query 2.5.28 (was 2.4.24) Perpendicular to x - 2 y = -5 containing (0,4) **** Give your equation for the requested line and explain how you obtained it.
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RESPONSE --> x - 2 y = -5 containing (0,4) -2y = -x - 5 y = 1/2x + 5/2 Slope for perpendicular line is: -2 y = -2x + 4
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17:27:40 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line perpendicular to this will therefore have slope -2/1 = -2. Point-slope form gives us y - 4 = -2 * (x - 0) or y = -2 x + 4. **
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RESPONSE --> ok
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17:29:43 **** query 2.3.10 (was 2.4.30). (0,1) and (2,3) on diameter **** What are the center, radius and equation of the indicated circle?
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RESPONSE --> r^2 = sqrt((0 - 2)^2 + (1 - 1)^2) r^2 = 2 (x - 2)^2 + (y - 1)^2 = 2 center is (2,1) Radius = 2
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17:29:55 ** The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = 2. This distance is a diameter so that the radius is 1/2 (2) = 1. The equation (x-h)^2 + (y-k)^2 = r^2 becomes (x-1)^2 + (y-2)^2 = r^2. Substituting the coordinates of the point (0, 1) we get (0-1)^2 + (1-2)^2 = r^2 so that r^2 = 2. Our equation is therefore (x-1)^2 + (y - 2)^2 = 2. You should double-check this solution by substituting the coordinates of the point (2, 3). **
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RESPONSE --> ok
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17:30:26 **** query 2.3.16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3?
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RESPONSE --> (x - 1)^2 + (y - 0)^2 = 3^2
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17:30:33 ** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r. In this example we have (h, k) = (1, 0). We therefore have (x-1)^2 +(y - 0)^2 = 3^2. This is the requested standard form. This form can be expanded and simplified to a general quadratic form. Expanding (x-1)^2 and squaring the 3 we get x^2 - 2x +1+y^2 = 9 x^2 - 2x + y^2 = 8. However this is not the standard form.
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RESPONSE --> ok
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17:33:59 query 2.3.24 (was 2.4.40). x^2 + (y-1)^2 = 1 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?
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RESPONSE --> x^2 + (y-1)^2 = 1 (x - 0)^2 + (y - 1)^2 = 1 x^2 + (0 - 1)^2 = 1 x = 0 x intercept (0,0) Center = (0,1) 0^2 + (y - 1) = 1 y - 1 = +-1 y - 1 =1 y = 2 (0,2) y - 1 = -1 y = 0 (0,0) y intercepts = (0,2) (0,0)
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17:34:05 ** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r. In this example the equation can be written as (x - 0)^2 + (y-1)^2 = 1 So h = 0 and k = 1, and r^2 = 1. The center of the is therefore (0, -1) and r = sqrt(r^2) = sqrt(1) = 1. The x intercept occurs when y = 0: x^2 + (y-1)^2 = 1. I fy = 0 we get x^2 + (0-1)^2 = 1, which simplifies to x^2 +1=1, or x^2=0 so that x = 0. The x intercept is therefore (0, 0). The y intercept occurs when x = 0 so we obtain 0 + (y-1)^2 = 1, which is just (y - 1)^2 = 1. It follow that (y-1) = +-1. If y - 1 = 1 we get y = 2; if y - 1 = -1 we get y -2. So the y-intercepts are (0,0) and (0,-2)
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RESPONSE --> ok
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17:39:32 **** query 2.3.34 (was 2.4.48). 2 x^2 + 2 y^2 + 8 x + 7 = 0 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?
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RESPONSE --> 2 x^2 + 2 y^2 + 8 x + 7 = 0 x^2 + 4x + y^2 = -7/2 x^2 + 4x + 4 + y^2 = 1/2 (x + 2)^2 + (y - 0)^2 = 1/2 (0 + 2)^2 + (y - 0)^2 = 1/2 4 + y^2 = 1/2 y^2 = -7/2 y^2 cannot be a negative number so there is no y-intercept (x + 2)^2 + (0 - 0)^2 = 1/2 x + 2 = sqrt(1/2) x = sqrt(1/2) - 2 x = -1.3 (-1.3,0) x + 2 = -sqrt(1/2) x = -sqrt(1/2) - 2 x = -2.7 (-2.7,0) x intercepts = (-1.3,0) (-2.7,0)
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17:39:38 ** Starting with 2x^2+ 2y^2 +8x+7=0 we group x and y terms to get 2x^2 +8x +2y^2 =-7. We then divide by the common factor 2 to get x^2 +4x + y^2 = -7/2. We complete the square on x^2 + 4x, adding to both sides (4/2)^2 = 4, to get x^2 + 4x + 4 + y^2 = -7/2 + 4. We factor the expression x^2 + 4x + 4 to obtain (x+2)^2 + y^2 = 1/2. From the standard form of the equation for a circle we see that the center is (-2,0) the radius is sqrt (1/2). To get the intercepts: We use (x+2)^2 + y^2 = 1/2 If y = 0 then we have (x+2)^2 + 0^2 = 1/2 (x+2)^2 = 1/2 (x+2) = +- sqrt(1/2) x + 2 = sqrt(1/2) yields x = sqrt(1/2) - 2 = -1.3 approx. x + 2 = -sqrt(1/2) yields x = -sqrt(1/2) - 2 = -2.7 approx If x = 0 we have (0+2)^2 + y^2 = 1/2 4 + y^2 = 1/2 y^2 = 1/2 - 4 = -7/2. y^2 cannot be negative so there is no y intercept. **
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RESPONSE --> ok
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17:42:17 **** query 2.3.30 (was 2.4.54). General equation if diameter contains (4, 3) and (0, 1). **** Give the general equation for your circle and explain how it was obtained.
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RESPONSE --> center = (4 + 0) / 2, (3 + 1)/ 2 = (2,2) r = sqrt((4 - 2)^2 + (3 - 2)^2) r = sqrt(5) (x - 2)^2 + (y - 2)^2 = sqrt(5)^2 (x - 2)^2 + (y - 2)^2 = 5
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17:42:22 ** The center of the circle is the midpoint between the two points, which is ((4+0)/2, (3+1)/2) = (2, 2). The radius of the circle is the distance from the center to either of the given points. The distance from (2, 2) to (0, 1) is sqrt( (2-0)^2 + (2-1)^2 ) = sqrt(5). The equation of the circle is therefore (x-2)^2 + (y-2)^2 = (sqrt(5))^2 or (x-2)^2 + (y-2)^2 = 5. **
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RESPONSE --> OK
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course Mth 158
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RESPONSE --> x - 2 y = -5 (0,0) -2y = -x - 5 y = 1/2x + 5/2 Slope for parallel line is 1/2 y = 1/2x
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17:26:22 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line parallel to this will therefore have slope 1/2. Point-slope form gives us y - 0 = 1/2 * (x - 0) or just y = 1/2 x. **
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RESPONSE --> ok
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17:27:36 **** query 2.5.28 (was 2.4.24) Perpendicular to x - 2 y = -5 containing (0,4) **** Give your equation for the requested line and explain how you obtained it.
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RESPONSE --> x - 2 y = -5 containing (0,4) -2y = -x - 5 y = 1/2x + 5/2 Slope for perpendicular line is: -2 y = -2x + 4
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17:27:40 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line perpendicular to this will therefore have slope -2/1 = -2. Point-slope form gives us y - 4 = -2 * (x - 0) or y = -2 x + 4. **
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RESPONSE --> ok
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17:29:43 **** query 2.3.10 (was 2.4.30). (0,1) and (2,3) on diameter **** What are the center, radius and equation of the indicated circle?
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RESPONSE --> r^2 = sqrt((0 - 2)^2 + (1 - 1)^2) r^2 = 2 (x - 2)^2 + (y - 1)^2 = 2 center is (2,1) Radius = 2
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17:29:55 ** The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = 2. This distance is a diameter so that the radius is 1/2 (2) = 1. The equation (x-h)^2 + (y-k)^2 = r^2 becomes (x-1)^2 + (y-2)^2 = r^2. Substituting the coordinates of the point (0, 1) we get (0-1)^2 + (1-2)^2 = r^2 so that r^2 = 2. Our equation is therefore (x-1)^2 + (y - 2)^2 = 2. You should double-check this solution by substituting the coordinates of the point (2, 3). **
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RESPONSE --> ok
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17:30:26 **** query 2.3.16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3?
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RESPONSE --> (x - 1)^2 + (y - 0)^2 = 3^2
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17:30:33 ** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r. In this example we have (h, k) = (1, 0). We therefore have (x-1)^2 +(y - 0)^2 = 3^2. This is the requested standard form. This form can be expanded and simplified to a general quadratic form. Expanding (x-1)^2 and squaring the 3 we get x^2 - 2x +1+y^2 = 9 x^2 - 2x + y^2 = 8. However this is not the standard form.
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RESPONSE --> ok
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17:33:59 query 2.3.24 (was 2.4.40). x^2 + (y-1)^2 = 1 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?
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RESPONSE --> x^2 + (y-1)^2 = 1 (x - 0)^2 + (y - 1)^2 = 1 x^2 + (0 - 1)^2 = 1 x = 0 x intercept (0,0) Center = (0,1) 0^2 + (y - 1) = 1 y - 1 = +-1 y - 1 =1 y = 2 (0,2) y - 1 = -1 y = 0 (0,0) y intercepts = (0,2) (0,0)
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17:34:05 ** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r. In this example the equation can be written as (x - 0)^2 + (y-1)^2 = 1 So h = 0 and k = 1, and r^2 = 1. The center of the is therefore (0, -1) and r = sqrt(r^2) = sqrt(1) = 1. The x intercept occurs when y = 0: x^2 + (y-1)^2 = 1. I fy = 0 we get x^2 + (0-1)^2 = 1, which simplifies to x^2 +1=1, or x^2=0 so that x = 0. The x intercept is therefore (0, 0). The y intercept occurs when x = 0 so we obtain 0 + (y-1)^2 = 1, which is just (y - 1)^2 = 1. It follow that (y-1) = +-1. If y - 1 = 1 we get y = 2; if y - 1 = -1 we get y -2. So the y-intercepts are (0,0) and (0,-2)
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RESPONSE --> ok
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17:39:32 **** query 2.3.34 (was 2.4.48). 2 x^2 + 2 y^2 + 8 x + 7 = 0 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?
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RESPONSE --> 2 x^2 + 2 y^2 + 8 x + 7 = 0 x^2 + 4x + y^2 = -7/2 x^2 + 4x + 4 + y^2 = 1/2 (x + 2)^2 + (y - 0)^2 = 1/2 (0 + 2)^2 + (y - 0)^2 = 1/2 4 + y^2 = 1/2 y^2 = -7/2 y^2 cannot be a negative number so there is no y-intercept (x + 2)^2 + (0 - 0)^2 = 1/2 x + 2 = sqrt(1/2) x = sqrt(1/2) - 2 x = -1.3 (-1.3,0) x + 2 = -sqrt(1/2) x = -sqrt(1/2) - 2 x = -2.7 (-2.7,0) x intercepts = (-1.3,0) (-2.7,0)
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17:39:38 ** Starting with 2x^2+ 2y^2 +8x+7=0 we group x and y terms to get 2x^2 +8x +2y^2 =-7. We then divide by the common factor 2 to get x^2 +4x + y^2 = -7/2. We complete the square on x^2 + 4x, adding to both sides (4/2)^2 = 4, to get x^2 + 4x + 4 + y^2 = -7/2 + 4. We factor the expression x^2 + 4x + 4 to obtain (x+2)^2 + y^2 = 1/2. From the standard form of the equation for a circle we see that the center is (-2,0) the radius is sqrt (1/2). To get the intercepts: We use (x+2)^2 + y^2 = 1/2 If y = 0 then we have (x+2)^2 + 0^2 = 1/2 (x+2)^2 = 1/2 (x+2) = +- sqrt(1/2) x + 2 = sqrt(1/2) yields x = sqrt(1/2) - 2 = -1.3 approx. x + 2 = -sqrt(1/2) yields x = -sqrt(1/2) - 2 = -2.7 approx If x = 0 we have (0+2)^2 + y^2 = 1/2 4 + y^2 = 1/2 y^2 = 1/2 - 4 = -7/2. y^2 cannot be negative so there is no y intercept. **
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RESPONSE --> ok
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17:42:17 **** query 2.3.30 (was 2.4.54). General equation if diameter contains (4, 3) and (0, 1). **** Give the general equation for your circle and explain how it was obtained.
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RESPONSE --> center = (4 + 0) / 2, (3 + 1)/ 2 = (2,2) r = sqrt((4 - 2)^2 + (3 - 2)^2) r = sqrt(5) (x - 2)^2 + (y - 2)^2 = sqrt(5)^2 (x - 2)^2 + (y - 2)^2 = 5
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17:42:22 ** The center of the circle is the midpoint between the two points, which is ((4+0)/2, (3+1)/2) = (2, 2). The radius of the circle is the distance from the center to either of the given points. The distance from (2, 2) to (0, 1) is sqrt( (2-0)^2 + (2-1)^2 ) = sqrt(5). The equation of the circle is therefore (x-2)^2 + (y-2)^2 = (sqrt(5))^2 or (x-2)^2 + (y-2)^2 = 5. **
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RESPONSE --> OK
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