the equation represents a sphere 4x^2 + 4y^2 +4z^2 -24x + 40y = 9... Find its center. enter your answer as an?
353472
2)(x-7)^2 + (y-6)^2 + (z-9)^2 = 36\n3)a)1 b) 3 c) 8\nFor 2 it is 36 because you use the absolute value of the smallest number of the center.  The smallest value is 6.\n6^2 = 36
869748
It's a circle of radius 4, centered at the origin.\n\nThe equation of a circle centered at (a, b) is:\n\n(x - a)² + (y - b)² = R²\n\nWhen the circle is centered at the origin, (a, b) = (0, 0), or:\n\nx² + y² = R²\n\nTherefore, y is simply given as:\n\ny = ±sqrt( R² - x² )\n\nIn this case, R² = 16, so R = 4. The plot you want to make is one such that every point on the plot is 4 units away from the origin.
851885
You have :\n | x+4y   =12\n | 2x-3y  =6\n<=>\n | 2x+8y=24  \n | 2x-3y=6              \n<=>  \n | 11y=18\n | x+4y=12\n<=>\n | y=18/11\n | x=12-4(18/11)=60/11\n\nFinally, we have (x=60/11) and (y=18/11) .\n ^_^
835073
okay well\n\n1) combine two to figure out the first variable\nx-2y+z=4\n-x+2y+3z=-4\nso 4z=0; z=0\nthen figure out the next variable with a different set of equations\nlike and multiply so the ys will cancel [take Zs out cause zero]\nx-2y+z=4\n2(2x+y)=6\n\ntherefore\nx-2y=4\n4x+2y=6\n5x=10\nx=2\n\nthen substitute for y\n\n-x+2y=-4\n-2+2y=-4\n2y=-2\ny=-1\n\nso y=-1, x=2, z=0\n\n\ncontinue the process with the rest.
1051508
Your equation would look like x^2-ax+b.  Which means when you solve the equation you would two answers one of which will contain a plus sign other will contain a minus sign since the middle term is negative.  Thus the answer is C.
817461
a^4 - b^4\n\n= (a^2 + b^2)(a^2 - b^2)\n\n= (a^2 + b^2)(a + b)(a - b)
986254
(z^4)^2 - 14z^4 +16 =0\nQuadratic factoring:\nz^4 = (14+sqrt(196-64))/2 = 7+sqrt(33) =\n12.7446 so z = 1.89 or -1.89\nOR\nz^4 = 7-sqrt(33)= 1.2554\nso z= 1.059 or -1.059
70526
=((-4y)^2)/(10y-5)= (16y^2)/(10y-5)\nand if the equation is equal to zero then y=4i
419498
First, group the terrms with x in parenthesis and the terms with y in parenthesis:\n(x^2 + 6x)  + (y^2 - 8y) = -9\nLet's think for a moment, we have two squared terms, x^2 and y^2. Think of the two groups in parenthesis as if you were completing the square to solve a quadratic equation.\nFor both, take 1/2 the middle term and square it, adding it to both sides.:\n(x^2 + 6x +(3)^2) + (y^2 - 8y + (4)^2) = -9 + (3)^2 + (4)^2\nFactor and combine like terms:\n(x + 3)^2 + (y - 4)^2 = 16\nNow we have the standard form of a circle\nThe center is the OPPOSITE of the numbers in parenthesis, according to PLNR (positive left, negative right)\nThus the center is (-3, 4) and radius is sqrt(16)= plus/minus 4\nRemember that the radius must extend in negative directions: down and left. That's why it is plus/minus.
844053
I'll rewrite the question for you, because we usually denote ^ as "to the power of" for simplicity.\n\nx^2 + y^2 - 4x + 2y = 4\n\nWe have to do a process called completing the square, for both the x and y variables.  So first, let's group all the x terms together and y terms together.\n\nx^2 - 4x + y^2 + 2y = 4\n\nTo complete the square, we have to add 4 to the x terms and 1 to the y terms.  Since we're adding 4 and 1 to the left hand side, we have to do the same to the right hand side.\n\nx^2 - 4x + 4 + y^2 + 2y + 1 = 4 + 4 + 1\n\nNow, we factor those now square polynomials.\n(x-2)^2 + (y+1)^2 = 9\n\nThe center of the circle is given by the values in the brackets, but the negative of them.  In the above case, the center is at (2,-1), since you take the negative of -2 (the value next to the x) and 1 (the value next to the y).\n\nTo get the radius, you take the square root of the right hand side, so the square root of 9 is 3.
1061975
if you have: 36x^2-4y^2=4\n\nsimplify by dividing every term by 4 and you get\n\n9x^2 - y^2 = 1\n\nthe general equation of a hyperbola is :\n\n[x^2 / a^2] - [y^2 / b^2] = 1\n\nif we rewrite 9x^2 - y^2 = 1 to look like the standard equation, we get\n\n[x^2 / (1/9)] - [y^2 / 1] = 1 \n\nwhere a = 1/3 and b = 1\n\nsince the x term is first, the foci are at (+c , 0) and (-c , 0)\nif the y term was first, the the foci will be at (0,+c) and (0,-c)\n\nto find c, we always use\n\nc^2 = a^2 + b^2, since i know what a and b are\n\nc^2 = (1/3)^1 + (1)^2 = 1/9 + 1 = 10/9\nc^2 = 10/9\ntake square root of both sides, c = +-1/3 rad10\n\nalso, since the x term is first, the vertices are at (+-a,0) = +-1/3,0\nif the y term was first, vertices would be at (0,+-a)\n\nlastly, for the asymptotes it also matters that the x term is first.\nasymptotes: y = (+-b / a)x = +-1/(1/3)x = +-3x\n(i think you made a mistake here)\n\nif the y term was first, the asymptotes would be\ny = (+-a / b)x
752254
Well, let's check if y=4.\nJust replace y with 4\n4(4) + 2(4) = 24\n16 + 8 = 24\n24 = 24 \nThis is equation is true, so you are correct.
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
835686
4x^2 + 3y^2 = 4\n8x x' + 6y y' = 0\n8 x '' + 6 y'' = 0\n6y'' = -8x ''\ny '' = -4x / 3
678883
(2y)(2y)+14y=-24 \n=> 4y^2+14y+24=0\n=>2y^2 + 7y + 12 =0 As 2 is never 0\n=> y = [-7 +/- sqrt(7^2-4*2*12)]/4\n=> y = -7/4 +/- sqrt(49-96)/4\n=> y = -7/4 +/- sqrt(-47)/4\nSo the complex root of y are:\n-1.75+1.713914i and\n-1.75-1.713914i
861681
subtract 1st equation from the second:\n  8y^2=32\nso y^2=4, and y = +/- 2\n\nplug back into 1st equation:\n4x^2+4=4\n\n4x^2=0\nx=0
281569
24y^3+56y^2-6y-14=\n\n24y^3-6y+56y^2-14=\n\n6y(4y^2-1)+14(4y^2-1)=\n\n(4y^2-1)(6y+14)=\n\n(2y-1)(2y+1)*2*(3y+7)=\n\n2(2y-1)*(2y+1)*(3y+7)