Can someone please show me the steps so I can graph this?\n\n(y+2)^2/25 + (x-3)^2/16 = 1
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First, I'll assume you mean that this is an ellipse (parenthesis are important):\n((y+2)^2)/25 + ((x-3)^2)/16 = 1\n\nTo graph by hand, start at the origin.  \n1)The center is shifted- to find it's location, set (y+2) = 0 and (x-3) = 0, and solve for x and y. The center is therefore at (-2,3). \n\n2)The half length of the axis (the lines bisecting the ellipse) are given by the denominators of the x and y terms.  To find the length of each axis, take the square root of each denominator.  The "radius" in the y (vertical) direction is sqrt(25) = 5 units long, and in the x direction it is sqrt(16) = 4 units long.  \n\n3)Starting at the center point of the ellipseee, (-2,3), count up half the length of the axis in the y direction, or 5 units, and put a point.  Do the same thing down, and then repeat with the x axis, counting left and right 4 units, and putting points on your graph.  Draw an ellips through these 4 points, and you're done! \n\nThe major axis is therefore the y axis, and is 10 units long, and then minor axis is the x axis, which is 8 units long.\n\nIf you want to graph this on a calculator, you must solve for y:\n((y+2)^2)/25 + ((x-3)^2)/16 -((x-3)^2)/16= 1 -((x-3)^2)/16\n\n(((y+2)^2)/25)*25 = (1 -((x-3)^2)/16)*25\n\n((y+2)^2)^(1/2) = ((1 -((x-3)^2)/16)*25)^(1/2)\n\nNote that raising to the (1/2) power is the same as taking the square root. When you take the square root of a real number, your answer could be negative or positive, so you have 2 solutions, one for the bottom of the ellips, and one for the top:\n\n(y+2) = (((1 -((x-3)^2)/16)*25)^(1/2))\nand\n(y+2) = -(((1 -((x-3)^2)/16)*25)^(1/2))\n\nThe final equations you'd enter into your calculator are:\n\ny = (((1 -((x-3)^2)/16)*25)^(1/2))-2\nand\ny = -(((1 -((x-3)^2)/16)*25)^(1/2))-2
554197
Pick 2 numbers for X, then plug them into your equation to get F(X), which is the Y value. Then you have the equations and put them onto your graph.\n\nTo plot the function, put your Y-intercept in (-3) then from there use the slope (-1/2) to plot other points. On the function, if you did it right you should come across your points that you made.\n\nHope this helps. Good Luck!!
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x+y=2\n2x-y=0\n\nx=2-y\n2(2-y)-y=0\n\nX=2-y     \n4-3y=0   \n\nx=2-4/3\ny=4/3
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b) Replace y with 0\n(100+x)^0 = 2*100^0\n1 = 2*1\n1 = 2\nThis implies that y cannot be zero.\nc) Replace x with 0.\n(100+0)^y = 2*100^y\n100^y = 2*100^y\n0 = 2*100^y - 100^y\n0 = 100^y\nThis is never true.\nThis implies that x cannot be zero.\na) The graph looks similar to the graph of y = 1/x
512691
y/9 = 250x^3 + 150 x^2 + 55x + 70\n\nTo get the x-intercepts, we set y=0:\n\n250x^3 + 150 x^2 + 55x + 70 = 0\n\n50x^3 + 30x^2 + 11x + 14 = 0\n\nLet f(x) = 50x^3 + 30x^2 + 11x + 14\n\nWe want to see where f(x)=0.  We note that f(0) = 14, and that for all positive x, f(x) > 0, so there are no positive roots.  Also, f(-1) = -50 + 30 - 11 + 14 = -17, so there's a root between -1 and 0.  Also, for all x < -1, f(x) < 0, so there are no roots less than -1.  All the roots are between -1 and 0.\n\nWe have two points, (0,14) and (-1,-17).  Let's try x = -0.5:\n\nf(-0.5) = 9.75  ==>  (-0.5, 9.75)  (I'm sort of interpolating)\n\nf(-0.7) = 3.85  ==>  (-0.7, 3.85)\n\nf(-0.8) = -1.2  ==>  (-0.8,-1.2)\n\nf(-0.78) = -0.06  ==>  (-0.78, -0.06)\n\nThat's pretty good, although we could go to a third decimal if we wanted to.  Just as a check, f(-0.785) = -0.34.  Let's go with one root at x = -0.78.\n\nNow let's divide that root out, reducing the cubic to a quadratic:\n\nf(x) = (x + 0.78)(50x^2 - 9x + 18) = 0\n\nThe roots of the quadratic g(x) = 50x^2 - 9x + 18 = 0 are imaginary.  Using the quadratic formula,\n\nx = 0.01 [9 +/- sqrt(81 - 3600)] = .01 [9 +/- 3 sqrt(9 - 400)]\n\nx = .03 (3 +/- i sqrt 391)\n\nso your only real root is x = -0.78.  (Answer)
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It's been a long time since I did these, but hopefully you understand how to graph:\n\nx-y=3    becomes   y=x-3\n\n3x-2y=3    becomes    y=1.5(x-1)\n\nNow graph these and use the definitions of systems in your textbook to determine which they are.
492940
y=1/x^2-3 => y+3 = 1/x^2 => x= +/-sqrt(1/(y+3))
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Look at the coefficient of the x^2 term.  If it is positive, the parabola opens up and has a minimum value; if it is negative, the parabola opens down and has a maximum.  You have 2x^2, so you will have a minimum value.\n\nNext find the vertex.  The x-coordinate of the vertex lies at -b/2a.  So -8/(2)(2) = -8/4 = -2.  Plug this value in to get the y-coordinate of the vertex.  2(-2^2) + (8)(-2) + 5 = 8 - 16 + 5 = -3.\n\nSo you have a minimum value of -3.
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y = 3/2 (-4) -2 = -12/2 - 2 = -6 -2 =  -8\n\n(x,y) = (4,-8)
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The x-coordinate of the vertex is centered between the zeros, so it is x = (-2 + 8)/2 or x = 3\n\nVertex form is y = a(x-h)^2 + k where (h,k) is the vertex.\n\nSo far we have y = a(x-3)^2 + k\n\nBut when x=0, y=8 since it's the y-intercept.\n\nSo 8 = a(0-3)^2 + k\n8 = 9a + k\n\nAnd when x is 8, y is 0 because 8 is a zero.\n\nSo 0 = a(8-3)^2 + k\n0 = 25a + k\n\nMultiply 8 = 9a + k by -1 and add it to 0 = 25a + k\n\nThis solves for a.  Finally plug this a value in to get k.
440278
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
525727
well to solve the first equation (its a line plot)\n3y+1[-1]=2x+16[-1]\n3y=2x+15\nif x=0, y=5 (y intercept)\nif y=0, x=-7.5 (x intercept)\n\nthe second line graph\n3y+5[+4]=2x\n3y+9=2x\nif x=0, y=-3\nif y=0, x=4.5\n\nNow, these lines are not equal, but do you have to plot both lines to prove? Your question is not clear.
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Equation #1 >>> x^2+y^2=1\n\nLet the value S=(1+x+iy)/(1+x-iy)\nmultiplying by the conjugate of the denominator (1+x+iy)\nS=[(1+x+iy)(1+x+iy)]/ [(1+x-iy)(1+x+iy)]\nS=[1+x+iy+x+x^2 +ixy+iy+ixy-y^2]/ [(1+x)^2-(iy)^2]\nnote:i^2=-1 and (x+y)(x-y)=x^2-y^2 but (x+iy)(x-iy)=x^2+y^2\n\nS=[1+x^2+2x+2iy+2ixy-y^2]/ [1+2x+x^2+y^2]\nsubstitut in the numerator 1-y^2=x^2 from eqn #1\nsubstitut in the denominator x^2+y^2=1 from eqn #1\n\nS=[2x^2+2x+2iy+2ixy]/[2+2x]\nDivide by 2 on the num.and den.\n\nS=[x(1+x)+iy(1+x)]/[1+x]\ndivide by (1+x) on the num and den.\n\nS=x+iy\n\nthis is the value required
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To graph functions, first get a piece of graph paper or you can free hand the X-Y axis and plot the points between -5 and 5 on both axis.\n\n1) h(x) = 2/x  domain of x (-1, 0, 1, 2)\n\nFind the coordinate pairs for each member of the domain.\nh(-1) = 2/-1  -->  (-1, -2)\nh(0) = 2/0  -->  undefined so no point on the graph\nh(1) = 2/1  -->  (1,2)\nh(2) = 2/2  -->  (2,1)\n\nThe graph will be the three points above.  It is not a line since the domain is limited to the integers listed.\n\n\n2) f(x) = x-1  domain is x/x is a real numbers (which excludes x=0 from the domain, but includes all other real numbers)\n\nSince the equation is in slope-intercept form, i.e.  y = mx + b\nthe slope of the line is m=1 and the intercept on the y-axis is b=-1\n\nYou can draw the line that way, or determine two points on the line like this:\nf(-1) = -1 - 1  -->  (-1,-2)\nf(1) = 1 - 1  -->  (1, 0)\n\nSince the domain is the real numbers except 0, a solid line is drawn to extend beyond both points.  Either way you draw the line, make an open point (ie. a circle) on the line where x=0.  Don't draw the line through the open point.  The line is terminated with arrows to indicate that it extend to infinity in both directions
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x = (y^2 + 2y -1) / 2\ndx=ydy+dy\n1=(y + 1 )dy/dx\ndy/dx=1/(y+1)\nd2y/dx2 = -1 / (y+1)^2
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1 - The easiest way to do this is solve both for at least 2 points, then  graph both lines.\n2x + y = 4\nSet x=0,  then y=4\nSet y=0, then x=2\nGraph the line through (0,4) and (2,0)\n\nx + y = 3\nx=0, so y=3\ny=0, so x=3\nGraph line through (0,3) and (3,0)\n\n2 - Same\nx + y = -2\nx=0, y= -2\ny=0, x= -2\nGraph line through (0, -2) and (-2,0)\n\n3 - Algebra expression is defined as  "one or more algebraic terms in a phrase. It can include variables, constants, and operating symbols, such as plus and minus signs. It's only a phrase, not the whole sentence, so it doesn't include an equal sign."\n\nSince 9 + w = c, includes an equal sign this is not an "expression", it is an equation.
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(x^2+9)-25 is the same as (x^2 - 16) since 9 - 25 = -16.\n(x^2 - 16) can be factored down to (x+4)(x-4)  If you multiply these out using the distributive property of multiplication (or FOIL as they call it in high school), you'll see it works out.\n\nSo now you have [(x+4)(x-4)]/(x-4)\n\n[(x-4)/(x-4)] = 1, leaving you with (x+4), which is your answer.