course Mth 158 I hadn't planned on doing assignment 19 tonight, but got in a work mood and did it anyway. It is at the end of #18. Sorry for sending you the same SEND file twice.
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21:47:25 **** 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 --> First I think I would find the slope of the line. I would subtract x from each side and divide each side by -2= y=1/2x+5/2, the slope of the line is 1/2. Then I would use the point slope formula y-y1=m(x-x1) y-0=1/2(x-0) the equation for the parallel line is y=1/2x
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21:47:36 ** 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 --> YES!!! Yippee!!
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21:52:09 **** 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 --> I would first subtract x from each side and then divide by -2= y=1/2x+5/2 The slope is equal to 1/2. The slope of the perpendicular line is the negative reciprocal= -2. Then I use the point slope form y-y1=m(x-x1)= y-4= -2(x-0) y-4= -2x= y=-2x+4
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21:52:17 ** 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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21:59:25 **** 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 radius of this circle is 3 units. I know that the formula for a circle is (x-h)^2+(y-k)^2=r^2. I am a bit at a loss on this.
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22:03:37 ** 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 --> Okay, I now understand how to get the radius. I also see how the substitutions for the formula of a circle work. Then after the initial substitutions, the coordinates (0,1) are entered (0-1)^2+(1-2)^2=r^2 and r^2=2. So the equation would be (x-1)^2+(y-2)^2=2. A little more complicated, but I will stick with it.
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22:05:54 **** 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-h)^2+(y-k)^2=r^2 (x-1)^2+(y-0)^2= 9 (x-1)^2- y^2=9
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22:06:10 ** 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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22:09:48 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 --> The center of this circle (I think) is (0,1). I came to this conclusion from the equation, where x^2 is by itself and the y-k part has -1. I believe the radius of the circle is 1, since the radius is squared and only 1 is the result of 1^2. The circle intercepts the y axis at 1 and 2.
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22:11:11 ** 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 --> Okay, I didn't know the origin counted as both x int and y int. Also, I messed up on the sign for the (y-k) and put positive one instead of negative. Other than that, I do understand and I got it.
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22:21:52 **** 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 --> This is in general form. I group the x variables and the y variables and put the constant on the right. I then complete the square on the x variable and the y variable. = I also factor out a 2= 2(x^2+4x+4)+2(y^2)= -7/2 (x+2)^2+ (y^2)=-7/2+4= -7/2+8/2=( 1/2)^2 The center of the circle is 2,0, I think the radius is 1/4 or else I have messed up tremendously.
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22:25:24 ** 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 --> I was good until I squared the radius instead of taking the sqrt. I know to substitute 0 for x to get the y int and substitute 0 for y to get x int., so (x+2)^2+y^2=1/2 If y=0, (x+2)^2= 1/2, x+2= plus or minus sqrt of 1/2 approximately -1.3 and -2.7 if using a calculator. if x=0, 4+y^2=1/2, y^2= 1/2-4= -7/2, but because y^2 can't be negative, then there is no y intercept.
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22:27:16 **** 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 --> The general equation of a circle is x^2+y^2+ax+by+c=0 Are you wanting the general form or the actual standard form? I was a little confused here,
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22:30:03 ** 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 --> Okay, this was the standard form then. So you would first find the center of the circle, which was the midpoint of the diameter (4+0)/2, (3+1)/2= (2,2) The radius is the distance from the center to one of the listed points. Distance from (2,2) to (0,1) is the sqrt (2-0)^2+(2-1)^2= sqrt5 The equation is (x-2)^2+(y-2)^2= sqrt of 5^2 (x-2)^2+(y-2)^2=5. I totally understand this, now that I understand the question.
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