1. Determine whether the given point is a solution of the equation.\n\n(-5, 1); 2p - 3q = -13\n\n2.Which quadrant does the ordered pair (0, 0) fall in?\n\n3.Which quadrant does the ordered pair (-4, 0) fall in?\n\n4.Graphs of linear equations create what type of lines?\n\n5.What is the name of this formula?\n\ny = mx + b\n\n6.Graphs that do not produce straight lines are the result of what type of equations?\n\n7.When considering the following equations, what type of graphs do these equations form?\n\ny = -1/2x + 3\n\ny = 1/x\n\ny = Ea1 - 5\n\ny = |x|\n\n\n8.Is the set of ordered pairs considered a function?\n\n{(-3, 2), (5, 2), (-3, 7), (1, 8)}\n \n9.What is the domain of this set of ordered pairs?\n\n{(-3, 2), (5, 2), (-6, 7), (1, 8)}\n\n10.What is the range of this set of ordered pairs?\n\n{(-3, 2), (5, 9), (-6, 7), (1, 8)}\n\n11.Name the parts of this equation\n\ny = mx + b\n\n12.Find the y-intercept for the equation below.\n\ny = 7x + 8\n\n13.Find the y-intercept in the equation below.\n\nf(x) = -6x - 2/3
208635
Seeing as you've gone to all that trouble writing it out, and I'm bored...\n\n1. 2(-5)-3(1)=-13, so yes it is correct.\n2. It is the origin, so it lies in all quadrants.\n3. Third and fourth quadrant.\n4. Straight lines, as the variables are of only the first degree.\n5. Basic line equation for a straight line, so linear. m is gradient, c is y-intercept.\n6. This depends on the degree of the variable, e.g. x^3 is a cubic function.\n7-10. Do yourself!\n11. m- gradient, x- explanatory variable, b- y-intercept.\n12. y intercept from above logic is 8, as b=8. Also, y intercepts line x=0. Let x=0 and y=8.\n13. Assuming f(x)=y, then again the answer is -2/3.\n\n\nThis really is basic!
38018
write the equation of the line containing the point (2,3) that is perpendicular to the line containing the points(4,6)and (0,0) \nThe slope of the line containing the points (4,6) and (0,0) is:\nm = (change in y)/(change in x)\nm = (y2 - y1)/(x2 - x1)\nm = (6 - 0)/(4 - 0)\nm = 6/4 = 3/2\nBeing perpendicular to a line means having the negative reciprocal of the slope of that line, so the slope of our line is:\nm = -2/3\nSince it passes to (2,3), we can now solve for the equation using the point-slope form of the line:\ny - y1 = m(x - x1)\ny - 3 = -(2/3)(x - 2)\n3(y - 3) = -2(x - 2)\n3y - 9 = -2x + 4\n2x + 3y = 4 + 9\n2x + 3y = 13 (Standard form)\ny = -(2/3)x + 13/3 (Slope-Intercept form)\n2x + 3y - 13 = 0 (General Form)\n\nHope this helps ;-)
431058
Hey, here's the setup for these types of problems.\nYou want the equation in the form of y = mx + b\nsince we know that x = 2, y = -7, and m = -1/5, we can plug these values into that equation to find the equation of the line that passes through the desired point. SO:\n-7 = (-1/5)(2) + b\n-7 = (-2/5) + b\nb = (-7)+(2/5)\nb = -33/5\nso the equation for the line is:\ny = (-1/5)x - (33/5)\n\nhope that helps!!
268221
Let the coordinates of point P be (x,y). Here, x and y represent general points.\n\n1) Find an expression for the slope of OP. To help you get started, I will tell you the answer directly. It is (y/x).\n\n2) Find an expression for the slope of AP. Try this yourself!\n\n3) Use the statement "slope of OP is 1 less than the slope of AP" to construct an equation in terms of y and x.\n\n4) Carry out some algebraic manipulation to obtain your final answer!\n\n5) Lastly, take a look at your final answer. Are there some coordinates that point P cannot take? For instance, your final answer should be an equation that has (0,0) as 1 of the solutions. But can point P take this set of coordinates? What other coordinates can P not take?
413538
Graph equations are of the form y=mx+c\n\nm is the slope or gradient\nc is the y intercept.\n\nSo, for y=-x+1, \nm= -1\nc = 1\n\nAnd for 2y=3x, you have to rearrange it to make it look like y=mx+c.\n\nSo divide by 2.\n\ny=3x/2\n\nm= 3/2 \nc =0\n\nHope you understood my explanation :)
402488
Get the equations in slope-intercept form (y=mx+b)\n\ny=1/2+1\ny=-2x-3/4\n\ny=-2/3+4\ny=-2/3x+8\n\nNow compare the slopes(m) of each equation\n\nif the slopes of the equations equal -1 (they are oppostite recipricals (example... 2 and -1/2)when multiplied they are perpendicular\n\nif the slopes are the same they are perpendicular \n\nif they are neither they are just intersecting lines
552156
1st one:\n\nadd the 2 equations\n(-4x + 4x) cancels\n(-5y + 4y) = -y\n(-21 + 16) = -5\n\ntherefore y = 5\nsubstitute y = 5 into one of the equation\n\n4x + 4(5) = 16\n4x + 20 = 16\n4x = 16 - 20\n4x = -4\nx = -1\n\n2nd one:\n\nmultiply the first equation by 2 since no addition can cancel x or y\n\n-4x + 2y = 28\nthe other equation remains the same : 4x - 4y = -36\n\nadd the 2 equations cancelling x\n\n2y - 4y = 28 - (-36)\n-2y = 64\ny = - 32\n\nsubstitute y = -32\n\n-2x + (-32) = 14\n-2x = 14 + 32\n2x = -46\nx = -23\n\nThe important step is the first one in which you decide which letter you want to cancel. you have to the find the easiest method. Good luck
385694
2. The equation is (y-y1)/(y1-y2)=(x-x1)/(x1-x2)\n=>(y+2)(-2-0)=(x+1)(-1-0)\n=>(y+2)(-2)=(x+1)(-1)\n=>-y-2=-2x-2\n=>-y+2x=0=> 2x-y=0 which is the required equn.\n3.The equation is (y-4)(4+4)=(x-0)/(0-4)\n=>y-4/8 =x/-4\n=>8x=-4y+16  or,8x+4y-16=0  o4 2x+y-4=0 is the reqd.equation\n4.the eqn. is (y+3)/(-3-3)= (x+2)/(-2-0)\n=>(y+3)/-6=(x+2)/-2\n=>-6x-12=-2y-6\n=>-8x+2y-6=0 or,-4x+y-4=0 which is the equired equation
517629
Please calm down. Now do you want the line between (2,0) and (-2,-2) The way I do it is to get the point vector solution of the line and eliminate t. The point is (2,0). The vector is <-2 - 2, -2 -0>. Just subtract the first coordinate from the second. So the vector is <-4, -2> \nNow set up the point vector equation describing the postion of the graph at time t.\nJust point + vector*t in component form.\nThe x-components: x = 2 - 4t \nThe ys: y = 0 - 2t\n\nThere, that is it. But some nasty professors don't like this form, so we'll eliminate t. We plug y = -2t into first equation and get x = 2 + 2y.\n2y = x - 2. y = x/2 -1.
445157
Part 2 is easy.  The function y = 1/(1+x) has an infinite discontinuity at x = -1.  Also, y(0) = 1, and for large x (positive and negative), the function approaches the x-axis as an asymptote.  For positive x, y(x) > 0, and for large negative x, y(x) < 0.  It's easy to draw the graph.\n\nIn this problem, we care about y(-3) = -1/2; y(-2) = -1; and y(1) = 1/2.  We want the parabola y = ax^2 + bx + c that passes through the points (-3, -1/2), (-2, -1), and (1, 1/2).  Since all three points must satisfy the quadratic equation, we can plug in values and write these three equations:\n\n(1)  For x = -3:  9a - 3b + c = -1/2\n(2)  For x = -2:  4a - 2b + c = -1\n(3)  For x = 1:  a + b + c = 1/2\n\nThe solution to this set of three linear equations in three unknowns is a = 1/4; b = 3/4; and c = -1/2.  (Details of the solution omitted.)\n\nThe quadratic is y = 1/4 x^2 + 3/4 x - 1/2.  This function passes through the three points we identified, and\n\n1/(1+x) < 1/4 x^2 + 3/4 x - 1/2 over the range specified in the problem.  (Actually, there's equality at three points.)\n\nThat inequality can also be written as\n\n4/(1+x) < x^2 + 3x - 2\n\nI'll answer the other part of the question as best I can in the morning.  Right now, it's time to sleep.
146297
Well, first, write it in standard form (y=mx+b).\n\ny = -3x + 0\n\nThe m in the equation is the slope of the line, and the b is the y-intercept. \n\nSlope = -3\ny-Intercept = 0\n\nFor another example, y = 5x + 3 would have a slope of 5 and a y-intercept of 3.\n\n\nHope this helped.
485557
1. x-y=9---------------------(1)\nx+y=7-----------------------(2)\nadding\n2x=16\ndividing by 2\nx=8\nsub in 1,\ny=-1\n\n2. 5x-4y=12--------(1)\n3x-4y=4-------------(2)\n(1)-(2)\n2x=8\ndividing by 2\nx=4\nsub in 1 \ny=2\n\n3. 3x+3y=6-----------(1)\n2x-y=1-----------------(2)*3\n6x-3y=3---------------(3)\n(1)+(3)\n9x=9\ndividing by 9\nx=1\nsub in (1)\ny=1\n\n4. 3x-5y=13-----------(1)*3\n4x+3y=17--------------(2)*5\n9x-15y=39\n20x+15y=85\nadding\n29x=124\ndividing by 29\nx=124/29\nsub and find y\n\n5. x-y=1---------(1)\n3x-y=10---------(2)\n(2)-(1)\n2x=9\ndividing by 2\nx=9/2\nsuby=7/2\n\n6. 4x=2y-7\nthe other equation is missing\nx=(2y-7)/4\n\n7. y=-2x-----------(1)\n5x+3y=-3---------(2)\nsub y=-2x in (2)\n5x+3(-2x)=-3\n5x-6x=-3\n-x=-3\nx=3\nsub\ny=-6\n\n8. x+y=6---------(1)\n-2x+y=-3---------(2)\nsubtracting\n3x=9\ndividing by 3\nx=3\nsub\ny=3\n\n9. 2x+y=45-----------(1)\n3x-y=5-----------------(2)\nadding\n5x=50\ndividing by 5\nx=10\nsub\ny=25
341901
y= -(2/3)x^2 + (2/3)x + 4\n\n\n(P): y = ax^2 + bx + c\n\nPoints (0;4)  (-2;0)  (3;0) are element of (P)\n-->  4 = 0a + 0b + c --> c = 4\n0 = 4a - 2b + c\n0 = 9a + 3b + c\n\n--> 5a + 5b = 0 --> b = -a\n--> 4a + 2a + 4 = 0 --> a = -2/3,  b=2/3
452445
To figure this out, you need to know that an ordered pair is in the form: (x, y). Just take each ordered pair and stick it into the equation:\na) 3(1) = 2(-1) -1 ... 3 = -3 (Not correct.)\nb) 3(-1) = 2(-1) - 1 ... -3 = -3 (Correct.)\nc) 3(-2) = 2(-5/2) - 1 ... -6 = -6 (Correct.)\nd) 3(0) = 2(-1/2) - 1 ... 0 = -2 (Not correct.)\n\nSo, the correct answers are (b) and (c).
661990
>< you really should read the answers for your other questions and learn... hope you know this afterwards!\n\n\n6x-y=-4\n2x+2y=15\n\nuse second equation to find out what y is\n\nx+y=7.5 (divide everything by 2)\ny=7.5-x\n\nsubstitue 7.5-x as y in the first equation, find x\n\n6x-(7.5-x)=-4\n6x-7.5+x=-4\n7x=3.5\nx=0.5\n\nput in value of x in an equation and find y\n\n6(0.5)-y=-4\n3-y=-4\n-y=-7\ny=7\n\ncheck answers (use equation that was not used to find y):\n2(0.5)+2(7)=15 :)\n\nx=0.5, y=7\n\n\nx+y=1\n2x-y=-2\n\nuse first equation to find y (in relation to x) --- can find x too, but you have to plug in answer into second equation, and y is easier to deal with in 2nd equation...\n\ny=1-x\n\nput 1-x as value of y in the second equation\n2x-(1-x)=-2\n2x-1+x=-2\n3x=-1\nx=-1/3\n\nput in value of x into an equation to find y\n(-1/3)+y=1\ny=4/3\n\ncheck answers:\n2(-1/3)-(4/3)=-2 :)\n\nThe answers are x=-1/3, y=4/3\n\n\n\nthird problem:\n5x-3y=-11\nx-2y=2\n\nuse second equation to find out value of x in relation to y\nx=2+2y (add 2y to both sides to isolate x)\n\nput in that value as x for the first equation:\n5(2+2y)-3y=-11\n10+10y-3y=-11\n7y=-21\ny=-3\n\nsubstitute value of y in x-2y=2 to find x\nx-2(-3)2\nx+6=2\nx=-4\n\ncheck answers\n5(-4)-3(-3)=-11 :) answers match up\n\nhope that helped you learn how to solve this type of algebra questions!
606711
Yes, there are 3 equations and 3 unknowns, but 1 of the equations isn't linear.  You will not get unique solutions when you plug into the 3rd equation.\n\nLets go ahead and do it.\n\nFrom 1) y = ar/c + b/c\nFrom 2) x = dr/f + e/f\n\nPlug into 3)\n\nr^2 = (dr/f + e/f)^2 + (ar/c + b/c)^2\n\n= d^2r^2/f^2 + 2dre/f^2 + e^2/f^2 + a^2r^2/c^2 + 2arb/c^2 + b^2/c^2\n\nLets clear fractions by multiplying through by f^2*c^2\nr^2f^2c^2=\nc^2d^2r^2 + 2c^2dre + c^2e^2 + f^2a^2r^2 + 2f^2arb +f^2b^2\n\nGathering like degree terms of r:\n\n(f^2c^2 - c^2d^2 - f^2a^2)*r^2 - (2c^2de + 2f^2ab)*r - c^2e^2 - f^2b^2\n=0.\n\nSo, it can be reduced to a 2nd degree polynomial in r.  This will only have real solutions if the discriminant >= 0.\n\nLastly, imagine what each of these things is.\n\n1) The equation of a plane in 3 D space.\n2) The equation of another plane in 3 D space.\n3) The equation of a sphere in 3 D space.\n\nThe intersection of 1 and 2 will be one of the following: empty (they are parrallel), a single line (nice solution), or they coincide.\n\nThe intersection of the sphere with the intersection of 1 and 2 will have solutions only if the line passes through the sphere.  It might not.  \n\nSo, if the disrimenent from that mess above is > 0, you get a nice line that passes through the sphere and there will be exactly 2 solutions.\n\nOh, and the answer to the question, 'When would you ever use this" is that it comes up in optimization and business problems when you want to find where various curves meet.
451125
1.  Since f(x) = x² + bx + c, and you know f(4) = 9 (since (4,9) is a point on the parabola, specifically the vertex), you can solve for f(4) = \n\n16 + 4b + c = 9, so 4b + c = -7\n\nSince (4,9) is the vertex, you know the tangent at this point is zero, i.e., f '(4) = 0.  (I hope this is a calculus HW problem so that you know what I am talking about!)\n\nSo get f '(x) = 2x + b\n\nSince f '(4) = 0, we have 2(4) + b = 0, so b = -8\n\nSince b = -8, and 4b + c = -7, we can now solve for c:\n\n4(-8) + c = -7, so c = 25\n\n----------\n\n2.  Not sure I can help here.  I took the first and second derivatives of h(x) to determine that the function has a local min at x=7/3 and a local max at x=-1, but h(7/3) ≠ -5 and h(-1) ≠ 2, so not sure where to go from here, sorry!\n\n----------\n\n3.  A parabola has a 2nd-degree polynomial equation, or y = ax² + bx + c.  We know two values, namely (4,4) and (2,8) so we can set up two equations like thus:\n\n4 = a(4²) + b(4) + c\n8 = a(2²) + b(2) + c\n\nwhich we can rewrite as:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n\nRight now we have 3 unknowns and only 2 equations, but we can get a third equation by taking the first derivative, or f '(x) = 2ax + b.  Since the vertex is at (4,4) we know that f '(4) = 0, and we can get a third equation:\n\n0 = 2a(4) + b = 8a + b\n\nSo now we have 3 equations in 3 unknowns and we can solve for a, b, and c:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n0 = 8a + b\n\nSubtract equation 1 from equation 2, then divide by 2 to get:  \n2 = -6a - b\n\nNow add equation 3 to this to get 2 = 2a, or a=1.\n\nNow let's back-substitute to get b and c:\n\n2 = -6(1) - b, so b = -8\n\n8 = 4a + 2b +c\n8 = 4(1) + 2(-8) + c, so c = 20\n\nSo the equation of our parabola is:\nf(x) = x² -8x + 20\n\n(Double check to make sure f(4)=4 and f(2)=8).