course mth158
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11:23:14 query 2.4.30 (was 2.3.24). Slope 4/3, point (-3,2) Give the three points you found on this line and explain how you obtained them.
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RESPONSE --> well we know that for ever 3 it runs to the right it well also rise by 4 so if we move to the right 3 places from -3 we get 0 and up 4 we get 4 so the first point would be (0,4) now over 3 and up 4 from there we get (3,8) and over 3 up 4 from there we get (6,12) so my three point are. (0,4)(3,8)(6,12)
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11:25:17 STUDENT SOLUTION: (-3,2) slope 4/3. Move 3 units in the x direction, 4 in the y direction to get ((-3+3), (2+4)), which simplifies to (0,6) (-3,2) slope 4/3 = -4/-3 so move -3 units in the x direction and -4 in the y direction to get ((-3-3), (2-4)) which simplifies to (-6,-2) From (0,6) with slope 4/3 we move 4 units in the y direction and 3 in the x direction to get ((0+3), (6+4)), which simplifies to (3,10). The three points I obtained are (-6,-2), (0,6), (3,10).
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RESPONSE --> I moved up from 0 instead of 2 on the first one so it made all my y answers wrong but I know what i done wrong.
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11:38:10 query 2.4.36 (was 2.3.30). Line thru (-1,1) and (2,2) **** Give the equation of the line and explain how you found the equation.
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RESPONSE --> we first find the slope (2-1)/(2-(-1)=1/3 so the slope is 1/3 now using point slope we get y-1=1/3(x-(-1))=y-1=1/3x+1/3=y=1/3x+4/3 so the equation of this line in slope intecept form is y=1/3x+4/3
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11:38:35 STUDENT SOLUTION: The slope is m = (y2 - y1) / (x2 - x1) = (2-1)/(2- -1) = 1/3. Point-slope form gives us y - y1 = m (x - x1); using m = 1/3 and (x1, y1) = (-1, 1) we get y-1=1/3(x+1), which can be solved for y to obtain y = 1/3 x + 4/3.
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RESPONSE --> o.k.
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11:58:48 **** query 2.4.46 (was 2.3.40). x-int -4, y-int 4 **** What is the equation of the line through the given points and how did you find the equation?
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RESPONSE --> with a x-int of -4 and y-int of 4 the two point are (-4,0)and (0,4) so we find the slope 4-0/0-(-4)=4/4=1 so the slope is 1 and we know the y-int if 4 we have the equation y=1x+4
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11:59:24 STUDENT SOLUTION: The two points are (0, 4) and (4, 0). The slope is therefore m=rise / run = (4-0)/(0+4) = 1. The slope-intercept form is therefore y = m x + b = 1 x + 4, simplifying to y=x+4.
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RESPONSE --> I did not simplify but O.k.
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12:01:53 **** query 2.4.56 (was 2.4.48). y = 2x + 1/2. **** What are the slope and the y-intercept of your line and how did you find them?
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RESPONSE --> The equation is in slope int form so the slope is 2 and the y-int is 1/2 because m=slope and b=y-int slope is 2 y-int is 1/2
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12:02:12 ** the y intercept occurs where x = 0, which happens when y = 2 (0) + 1/2 or y = 1/2. So the y-intercept is (0, 1/2). The slope is m = 2.**
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RESPONSE --> o.k
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L????????????assignment #019 ???{?]???????? College Algebra 10-27-2005
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12:54:45 **** query 2.5.22 (was 2.4.18) Parallel to x - 2 y = -5 containing (0,0) **** Give your equation for the requested line and explain how you obtained it.
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RESPONSE --> 1/2x-y=0 to find this equation put the original equation into slope intercept form and got y=1/2x+5/2 so the slope was 1/2 I then took the point (0,0) and the slope and put them in this equation y-0=1/2(x-0)=y=1/2x or 1/2x-y=0
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12:55:29 ** 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 --> I did not leave it in slope intercept form but got the same answer.
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13:12:29 **** 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 --> y=-2x+4 is the equation perpindicular to the line x-2y=-5 we find this by first putting x-2y=-5 in slope int form -2y=-x-5=y=1/2x+5/2 so the slope is 1/2 the slope of the perpindicular line is the negative reciprocal of this slope it would be -2 we then take the point(0,4) and put it into point slope form and get y-4=-2(x-0) and solve y-4=-2x=y=-2x+4 so the equation we are looking for is y=-2x+4
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13:12:40 ** 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 --> o.k.
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11:38:18 **** 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 --> the center is the midpoint between (0,1)and(2,3) (0+2/2),(1+3/2)=(1,2)so the center of this circle is (1,2) Radius is distance from (1,2)to (2,3) =sqrt(2-1)^2+(2-3)^2 =sqrt1+1 =1.414 sso the radius of this circle is 1.414 the equation would then be (x-1)^2+(y-2)^2=2
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11:39:07 ** 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 --> o.k.
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11:45:47 **** 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 --> standard form is (x-h)^2+(y-k)^2=r^2 so the equation for this problem would be (x-1)^2+(y-0)^2=9
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11:46:30 ** 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 --> o.k.
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