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April 18, 2014

# Search: 6x + 2y + 8

Number of results: 10,607

Maths ..Equations
Isn't the 3-2y supposed to be in parentheses? Otherwise, 2y/6 should have been written y/3. If x = (3 -2y)/6, then 6x = 3 - 2y 2y = 3 - 6x y = (3/2) -3x
Monday, December 10, 2007 at 1:03pm by drwls

Math
Please help!!!! Solve each system graphically. Be sure to check your solution. If a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this. 6x - 2y = 2, 9x - 3y = 1 My answer: Equation 1 X-...
Monday, October 11, 2010 at 7:17pm by Jordyn

Surely you mean f(x) = (2x+8)/(x+6) and g(x) = (6x-8)/(2-x) f(g(x)) = (2(6x-8)/(2-x) + 8)/((6x-8)/(2-x) + 6) multiply by (2-x)/(2-x) = (2(6x+8) + 8(2-x))/(6x-8 + 6(2-x)) = (12x-16 + 16 - 8x)/(6x-8 + 12 - 6x) = 4x/4 = x check: let x=4 g(x) = 16/-2 = -8 then f(g(4)) = f(-8) = (-...
Wednesday, January 9, 2013 at 6:01pm by Reiny

MaTh = Algebra = help
x - 2y = 10 3x - y = 0 eliminate y by multiplying 3x - y = 0 by -2 -2(3x - y = 0) = -6x + 2y = 0 ADD the 2 equations x - 2y = 10 -6x + 2y = 0 -5x + 0 = 10 -5x = 10 x = -2 substitute x = -2 in x - 2y = 10, to find y x - 2y = 10 -2 - 2y = 10 -2y = 12 y = -6 check x = -2, y = -6 ...
Monday, January 17, 2011 at 8:51pm by helper

I need help simplifying this expression: (3x^3y^5 + 6x^2y^-2/6x^5y^-2)^2 (3x^3y^5 + 6x^2y^-2/6x^5y^-2)^2 the second term has a y^-2 in the numerator and denominator, so.. (3x^3y^5 + 6x^2 /6x^5 )^2 (3x^3y^5 + 1 /x^3 )^2 I am not certain much more can be done to simplify it.
Friday, May 18, 2007 at 12:46am by Courtney W.

It is helpful to use parentheses. Otherwise, what you're typing is incomprehensible. (6x^2)(2y^2) + (9x^2)(y^2)/(3x^2)(2y^2) Is this correct? Or is it this... [(6x^2)(2y^2) + (9x^2)(y^2)]/[(3x^2)(2y^2)] Please clarify your expression.
Tuesday, November 27, 2007 at 7:03pm by Michael

algebra
x^2+xy-2y^2 = (x+2y)(x-y) x^2+3xy-4y^2 = (x+4y)(x-y) so, the LCD = (x+2y)(x+4y)(x-y) and putting it all over that LCD, the numerator then becomes (6x-2y)(x+4y) - (3x+2y)(x+2y) = 6x^2 +22xy - 8y^2 - 3x^2 +8xy + 4y^2 = 3x^2 + 14xy - 12y^2 so, the final fraction is 3x^2 + 14xy - ...
Wednesday, April 25, 2012 at 2:31pm by Steve

math
I interpreted your statement as (2x^2)(y)(3xy^2 - 2y) - 3xy^2((2x^2)(y) - 5x) =6x^3y^3 - 4x^2y^2 - 6x^3y^3 + 15x^2y^2 = 11x^2y^2
Tuesday, August 26, 2008 at 11:10pm by Reiny

algebra
Hi J, 2y-6x=8 2y=6x+8 y=3x+4 In line 1, you are adding +6x to both sides of equation to get line 2 then inorder to get the value of y, (since you have 2y), divide both sides of the equation by 2. this will give your line #3 Hope this helps!
Wednesday, January 5, 2011 at 10:48pm by Kay

math
using implicit differentiation, 6x + 2y + 2xy' + 2yy' = 0 when x=1, y=-1, so plug them i: 6 - 2 + 2y' - 2y' = 0 Note that there is no solution, so the tangent to the curve is vertical. y' is not defined. Looking at dx/dy, 6xx' + 2x + 2yx' + 2y = 0 6x' + 2 - 2x' - 2 = 0 4x' = 0...
Sunday, November 4, 2012 at 8:04pm by Steve

Precalc
Write the equation of the hyperbola whose center is at the origin and has a vertical transverse axis. The equations of the asymptotes are 6x + 2y = 0 and 6x - 2y = 0. Is this correct? (y^2/9)-(x^2/4)=1
Sunday, August 21, 2011 at 4:17pm by Sara

Calculus
x^2 + y^2 - 6x - 2y + 4 = 0 x^2 - 6x + 9 + y^2 - 2y + 1 = -4 + 9 + 1 (x-3)^2 + (y-1)^2 = 6 I will leave it up to you to take it from there.
Thursday, July 5, 2012 at 11:08pm by Reiny

PreCalculus
I have to write the equation of a hyperbola whose center is (0,0) and which has a vertical transverse axis. The equations of the asymptotes are 6x+2y=0 and 6x-2y=0. I don't even know where to begin. Could someone help? Thanks.
Thursday, March 26, 2009 at 3:08pm by Joanie

algebra
2y-6x=8 2y=6x+8 y=3x+4 please explain this to me
Wednesday, January 5, 2011 at 10:48pm by j

Math
3x-5y=-11 (A) 6x=2y=26 (B) (A)*(2)6x-10y=-22(C) (B)*(5)30x+10=130(D) (C)+(D)36x =108 108/36=3 x=3 substitute in (A) or (B) choose easiest (B)- 18+2y=26 (take 18 from both sides 2y=8 8/2 y=4 so x=3 and y=4 (3,4)
Saturday, June 20, 2009 at 6:59pm by Nabby

Maths ..Equations
Hi I have to rearrnge the following to make y the subject.. x = 3 - 2y / 6 I get the first part do x 3, - 2, and / 6. Undo x 6, + 2, and / 3 then multiply by 6 so, 6x = 3 - 2y add 2 to both sides, 6x + 2 = 3 the divide both sides by 3 6x + 2 = y ------- 3 so y = 6x - 2...
Monday, December 10, 2007 at 1:03pm by Leo

Math
Walker's answer is very very wrong! You cannot add unlike terms. this is done by grouping 6x2y3 +18xy + 3x2y2 + 9x = 6x^2y^3 + 3x^2y^2 + 18xy + 9x = 3x^2y^2(2y + 3) + 9x(2y + 3) = (2y+3)(3x^2y^2 + 9x) = x(2y+3)(3xy^2 + 9)
Tuesday, March 4, 2008 at 9:42am by Reiny

algebra
2x+10y=92 -6x-2y=-24 multipy the first equation by 3 then add the equations. 6x-6x+30y-2y=276-24 28y=254 y= 254/28 = 127/14 then put that into either equation and solve for x
Tuesday, August 30, 2011 at 8:38pm by bobpursley

maths
easiest way: since it is parallel it must differ only in the constant let the equation be 2y - 6x = x at (3,-1) -2 - 18 = c c = -20 2y - 6x = -20 or 3x - y = 10
Monday, August 5, 2013 at 2:11pm by Reiny

implicit differentiation
With y considered a function of x, differentiate both sides of the equation with respect to x. Then solve for dy/dx. 2y dy/dx = (2x^3) dy/dx + y*(6x^2)+ 5 dy/dx(2y-2x^3) = 6x^2*y +5 dy/dx = [6x^2*y+5)/(2y-2x^3)
Saturday, May 17, 2008 at 9:34am by drwls

no it like this 6x^2y^3+9x^2y^3 ________________ 3x^2y^2
Tuesday, November 27, 2007 at 7:03pm by sharon

algebra
(6x-2y)/(x^2+xy-2y^2)-(3x+2y)/(x^2+3xy-4y^2)
Wednesday, April 25, 2012 at 2:31pm by Jenna

algebra
(x+2y)(x^2 + 6 xy -5 y^2) = x(x^2 + 6 xy -5 y^2)+2y(x^2 + 6 xy -5 y^2) = x^3+6x^2y-5xy^2+2x^2y+12xy^2-10y^3 = x^3 + 8x^2y +7xy^2 -10y^3 square inches
Sunday, February 6, 2011 at 2:30pm by Damon

algebra
I need to simplify (6x^2y^3)(9x^2y^3)/3x^2y^2. I don't know whether I'm supposed to multiply or divide or both?
Thursday, June 9, 2011 at 1:38pm by Lori

math
the simplest way for this example is to use substitution. sub x = -2y into 3x+4y = -2 3(-2y) + 4y = -2 -6x + 4y + -2 -2y = -2 y = 1 then since x = -2y x = -2(1) = -2
Tuesday, January 13, 2009 at 6:31pm by Reiny

Intermediate Algebra
A = (x+2y)(x^2+6xy-5y^2) A=x^3+6x^2y-5xy^2 + 2x^2y+12xy^2-10y^3 A = x^3 + 8x^2y +7xy^2 - 10y^3 A = x^2(x+8y) + y^2(7x-10y). Sq. In.
Friday, June 8, 2012 at 2:11am by Henry

algebra
4x+y=1 y=1-4x 6x+2y= -2 6x+2*(1-4x)= -2 6x+2-8x= -2 6x-8x= -2-2 -2x= -4 Divide with -2 x= -4/-2 x=2 y=1-4x y=1-4*2 y=1-8 y= -7 Proof: 6x+2y= -2 6*2+2*(-7)= -2 12+(-14)= -2 12-14= -2 -2= -2 4x+y=1 4*2+(-7)=1 8-7=1 1=1
Wednesday, April 13, 2011 at 5:25pm by Bosnian

implicit differentiation HELP
a web search for folium descartes tangent line reveals that the tangent is horizontal at x = 0 x = 4^(1/3) = 1.587 so, I was off somewhere. At one site, they work through without parametric equations: horizontal tangents occur when dy/dx = 0 vertical tangents occur when dx/dy...
Friday, October 28, 2011 at 3:11pm by Steve

Math
There is no solution, since the two equations correspond to parallel lines that do not intersect. In other words, they are incompatible equations. The second equation can be rewritten 6x - 2y = 2/3, which clearly disagrees with 6x - 2y = 2. I don't know what you mean by "set ...
Friday, August 5, 2011 at 9:03pm by drwls

intermediate algebra
Assuming you meant 6x^3yz^4 - 3x^2y^3z^4 - 18x^4y^2z^3, 3x^2yz^3(2xz- y^2z - 6x^2y)
Tuesday, July 26, 2011 at 11:23pm by PsyDAG

math
2x+2y=1 6x-2y=12 -4x=-11 x= 11/4 2(11/4)+2y=1 1-22/4=2y -18/4=2y -18/8=y
Saturday, April 7, 2012 at 9:02pm by Anonymous

Mathematics - Trigonometric Identities
Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y)(sin^2y) / (sin^...
Thursday, November 8, 2007 at 6:17pm by Anonymous

math
= x + 2y + 6x^2-12xy+8x = 6x^2 +9x + 2y -12xy
Thursday, July 18, 2013 at 1:55pm by Chol

Math Algr.2
the best way to avoid fractions is to multiply through by the GCD. #1. y = 3/2 x + 6 2y = 3x + 12 -3x + 2y = 12 or 3x-2y = -12 #2. -9y-6x-18 = 0 9y+6x+18 = 0 9y = -6x-18 y = -6/9 x - 18/9 y = -2/3 x - 2 A hood way to check is to pick a value for x, evaluate y, and see whether ...
Tuesday, November 6, 2012 at 6:12pm by Steve

algebra
what is a quartic function with only the two real zeros given x=5 and x=1 y=-x^4-6x^3+6x^2-6x+5 y=x^4+6x^3-6x^2-5 y=x^4-6x^3+5x^2-6x+6 y=x^4-6x^3+6x^2-6x+5 can someone explain
Wednesday, November 14, 2012 at 7:10pm by lee

MaThS
A. Y = -2x+1, and Y = -2x+3. m1 = m2 = -2. The lines are parallel, because their slopes are equal. B. x+y/3+5 = 0, and 2y+6x = 1 x+y/3=-5, and 6x+2y = 1. 3x+y = -15, and 6x+2y = 1 m1 = -A/B = -3/1 = -3 m2 = -6/2 = -3 The slopes are equal; therefore, the lines are parallel.
Sunday, August 11, 2013 at 12:23pm by Henry

calculus
5x^2-6y^2-30x-12y+9=0 lets complete the square 5(x^2 - 6x + ___) -6(y^2 + 2y + ___) = -9 5(x^2 - 6x + 9) -6(y^2 + 2y + 1) = -9 + 45 - 6 5(x-3)^2 - 6(y+1)^2 = 30 divide both sides by 30 (x-3^2 /6 - (y+1)^2 /5 = 1 I will leave it up to you to identify the conic and its properties.
Sunday, May 25, 2008 at 10:09pm by Reiny

multivariable calculus
If the three points are u,v,w then (v-u)x(w-u) = | x y z | | -1 2 0 | | 1 1 3 | = 6x+2y-3z plug in any of the points and you have 6x+2y-3z = 6
Tuesday, March 26, 2013 at 8:31am by Steve

calculus
3x^2+2xy+y2=2 6x + 2y + 2xy' + 2yy' = 0 x=1 ==> y=-1 6(1) + 2(-1) + 2(1)y' + 2(-1)y' = 0 6 - 2 + 2y' - 2y' = 0 4 = 0 (e) undefined The graph is an ellipse. At (1,-1) there is a vertical tangent
Sunday, February 26, 2012 at 4:06pm by Steve

someone please show how to do Simplify, if possible: 6x^2y^3+9x^2y^3/3x^2y^2
Tuesday, November 27, 2007 at 7:03pm by sharon

algebra
can somebody help me out? am I doing this right? using the elimination method. y = 3x + 2 2y-6x=4 We have the solution for y so we use this to solve 2 (3x + 2) – 6x = 4 6x + 4 – 6x = 4 0x +4 = 4 Subtract 4 from both sides X=0 Now we use 0 for the x value in equation one Y=3(0...
Saturday, July 30, 2011 at 9:17pm by kelly

Math
The way you typed it .... 3/2y-3/2y+4 = 4 If you meant .... 3/(2y-3) - 3/(2y+4) then = (3(2y+4) - 3(2y-3))/((2y-3)(2y+4)) = (6y+12-6y+9)/((2y-3)(2y+4)) = 21/((2y-3)(2y+4))
Sunday, May 16, 2010 at 4:49am by Reiny

Mathematics - Trigonometric Identities
I meant the reciprocals of SOH CAH TOA And ... I just got it Here's my answer: LS: = 1 + 1/tan^2y = 1 + 1/(sin^2y/cos^2y) = 1 + 1(cos^2y/sin^2y) = 1 + cos^2y / sin^2y = 1 + 1-sin^2y / sin^2y = 1(sin^2y) / sin^2y + 1 - sin^2y / sin^2y = sin^2y + 1 - sin^2y / sin^2y = 1/sin^2y
Thursday, November 8, 2007 at 6:17pm by Anonymous

Algebra 2
3x+y=5 multiply -2, your way 6x-5y=-4 to get -6x-2y=-10 6x-5y=-4 add to get -7y=-14 y=2
Saturday, November 13, 2010 at 7:16pm by bobpursley

Algebra 1
5x-2y=4 3x+y=9 5x -2y = 4 6x + 2y= 18 11x = 22 x = 2 y = 3 (2, 3)
Thursday, December 19, 2013 at 2:09pm by Kuai

Math
completing the square will give us the centre .... x^2 - 6x + .... + y^2 + 2y + ... = 15 x^2 - 6x + 9 + y^2 + 2y + 1 = 15 + 9+1 (x-3)^2 + (y+1)^2 = 25 centre is (3,-1) so you want the line through (3,-1) and (8,-2) slope = (-2+1)/(8-3) = -1/5 y+1 = (-1/5(x-3) 5y + 5 = -x + 3 x...
Thursday, December 1, 2011 at 11:38pm by Reiny

ALGEBRA
Do I place the following two equations on the same line as they both represent the same lines. Pleas show. Thanks much. 3x-2y=8 6x+4y=16 I am sorry I am not too big on graphing. no they are not the same line. because if you had reduced 6x+4y=16---> you would get 3x+2y=8 ...
Monday, January 8, 2007 at 3:05pm by DANIELLE

Algebra 1
Which of the following is parallel to the line: 4x+2y=-3? a) 4x+3=2y+3 c) 1/4x+1/2y=1/3 b) 6x+3y=-3 d) 2x-4y=6
Tuesday, May 24, 2011 at 11:10pm by Alfredo

math
1st equation ---> 2x + 2y = 1 2nd equation times 2 ---> 6x - 2y = 12 add them 8x = 13 x = 13/8 sub into 1st: 2(13/8+ 2y = 1 2y = 1 - 13/4 2y = -9/4 y = -9/8 x = 13/8 , y = -9/8
Saturday, April 7, 2012 at 9:02pm by Reiny

implicit differentiation
2y(dy/dx) = 2x^3(dy/dx) + (6x^2)y + 5 2y(dy/dx) - 2x^3(dy/dx) = (6x^2)y + 5 dy/dx(2y - 2x^3) = 6x^2y + 5 dy/dx = ((6x^2)y + 5)/(2y - 2x^3)
Saturday, May 17, 2008 at 9:34am by Reiny

Algerbra
LCD = 2y 3/(2y) + 4/(2y) - 2y/(2y) -4/(2y) = (3-2y)/(2y) or do you mean 3/(2y) +2/(y-1) -2/y ??? in that case LCD = [2y(y-1)] 3(y-1)/[2y(y-1)] +4y/[2y(y-1)]-4(y-1)/[2y(y-1)] or [3(y-1)+4y -4(y-1)]/[2y(y-1)] [3y+1]/[2y^2-2]
Saturday, November 24, 2012 at 10:17pm by Damon

Math
is this right? (x-5)^2+(y+1)^2=(x-3)^2 so -10x+25+y^2+2y+1=-6x+9 4x=y^2+2y+17
Tuesday, September 30, 2008 at 5:51pm by Anonymous

Substution Soluton MATH
1. since we have two expressions each equal to y, just equate them: 3x-14 = -5x+2 8x = 16 x=2 sub into one of them: y = 3x-14 = 6-14 = -8 2. from the first : y = -x into the 2nd: 5x -x = 4 4x=4 x = 1 then y = -1 3. probably a typo , you meant: 4x + 2y = 8 6x+2y=4 4x+2y=8 ...
Thursday, October 31, 2013 at 5:58am by Reiny

Math
Use the sign of the larger. 5x+4y-11x-2y -6x+2y
Saturday, September 12, 2009 at 12:34pm by bobpursley

algebra
4(2x-y)+2(y+3x) 8x - 4y + 2y + 6x 14x - 2y
Monday, September 5, 2011 at 11:36am by Ms. Sue

math
You should have written the "problem" with parentheses as 3/(2y) -3/(2y+4) . The way you wrote it, it is not clear what the denomonators are. You still have not asked a question. You cannot solve for y since it is not an equation. If you are supposed to rewite it with a common...
Thursday, May 20, 2010 at 12:27am by drwls

Precalc
Oops! An error on my part! The last line should be: =((2y+7)^5)x((y-2)^4)x[5x(2y+7)+6x(y-2)x2]
Tuesday, July 22, 2008 at 6:51pm by Quidditch

algebra 2
how to find equation of horizontal tangent line of 2y^3+6x^2y-12x^2+6y=1
Monday, April 4, 2011 at 1:24am by Anonymous

algebra
Simplify 5x^7y^6/ 4y^6x^3*7x^3y^4/6x^2y^5
Tuesday, February 3, 2009 at 10:39pm by kate

algebra
Simplify 5x^7y^6/ 4y^6x^3 * 7x^3y^4/6x^2y^5
Tuesday, February 3, 2009 at 10:39pm by kate

Math
(3x^2 - 12y^2) / (x^2 + 4xy + 4y^2), 3(x^2 - 4y^2) / (x + 2y) (x + 2y), 3(x + 2y) (x - 2y) / (x + 2y)(x - 2y), Cancel two (x + 2y) factors and get: 3(x - 2y) / (x + 2y).
Wednesday, October 13, 2010 at 6:05pm by Henry

Calculus Help and Check
x^5(x + y) = y^2(9x − y) 5x^4(x+y) + x^5(1+y') = 2yy'(9x-y) + y^2(9-y') 5x^5+5x^4y + x^5 + x^5y' = 18xyy' - 2y^2y' + 9y^2 + y^2y' y'(x^5 - 18xy - 3y^2) = 9y^2 -6x^5 - 5x^4y y' = (9y^2-6x^5-5x^4y)/(x^5-18xy-3y^2) you were close, but must have messed up some signs somewhere.
Friday, March 14, 2014 at 3:53pm by Steve

Math Homework
5x + 2y = -17 x = 3y 5(3y) + 2y = -17 15y + 2y = -17 17y = -17 y = -1 x = 3(-1) x = -3 (-3, -1) 2.) Solve the system by substitution. x/3 + y = 4/3 -x + 2y = 11 y = -x/3 + 4/3 -x + 2(-x/3 + 4/3) = 11 -x -2x/3 + 8/3 = 11 -5x/3 = =25/3 -3/5(-5x/3) = 25/3(-3/5) x = -5 (-5, 3) 3...
Saturday, December 7, 2013 at 1:47pm by Kuai

math
(X+2Y)+6X(X-2Y)+8X
Thursday, July 18, 2013 at 1:55pm by casey

Optimization (Math)
The mayor of a village wants to build a library of which the windows have a shape of a rectangle on top of a square. The total perimeter of each window is of P meters and varies depending on each windows size. Find the dimensions of the windows in terms of P, that maximizes ...
Tuesday, November 1, 2011 at 12:08pm by Tommy

Algebra
4x+2y-3z=7 2x+4y+z=35 6x+2y-2z=20 solve the system
Sunday, June 19, 2011 at 12:58am by Jack

In the following system of equations, what does x equal? -10x - 2y = 56 6x - 2y = -40 x = 6 or x = -6 or x = 2 or x = -2
Wednesday, April 24, 2013 at 9:45am by gage

algebra 1
-2x + 5y = -1 3x + 2y = 11 -6x + 15y = -3 6x + 4y = 22 19y = 19 y=1 so, x=3
Tuesday, January 17, 2012 at 10:19pm by Steve

Math
Did you mean 2(x^2)y 12 + 12xy + 18y ? If so, then = 2y(x^2 + 6x + 9) = 2y(x+3)^2 If you meant it the way you typed it, then use Damon's solution.
Monday, January 10, 2011 at 9:05am by Reiny

math
I disagree with all 3 of your answers 1. (4y-2)/(8y-2) = 2(2y-1)/((2(4y-1)) = (2y-1)/(4y-1) 2. x-3)/3x-5)/3-x/6x-10 this is not clear, do you mean (x-3/3x-5)/((3-x)(6x-10)) ? then (x-3)/(3x-5)*(6x-10/(3-x) = (x-3)/(3x-5)*2(3x-5)/(-x) = -2 , x not equal to 3 and 5/3 3. 3a+6/2a^...
Tuesday, March 24, 2009 at 8:16pm by Reiny

Geometry
Complete the squares for x and y to put into standard (x - h)^2 + (y - k)^2 = r^2 for center (h , k) and rdius r x^2 + 6x + y^2 - 2y = 15 x^2 + 6x + 9 + y^2 - 2y + 1 = 15 + 9 + 1 = 25 (x + 3)^2 + (y - 1)^2 = 25 comparing to the standard form: center (h , k) is (-3 , 1) r^2 = ...
Tuesday, May 3, 2011 at 4:26pm by Anonymous

Math
How do you factor "0 = x^4 - 6x^2 + 5"? Please show all steps. Here's my work, but I just went around in circles..... 0 = x^4 - 6x^2 + 5 -5 = x^4 - 6x^2 -5 = x^2 (x^2 - 6) -5 / x^2 = x^2 - 6 (-5 / x^2) + 6 = x^2 (-5 + 6x^2) / (x^2) = x^2 -5 + 6x^2 = x^2 * x^2 -5 + 6x^2 = x^4 0...
Thursday, May 22, 2008 at 10:06pm by Anonymous

Math
Solve each system using the method of your choice. 1. 7x+2y=17 2. 6x-5y=-9 3. y=4x-8 4. y=2x+10 5. 3x-6y=30 6. y=-6x+34 7. 3x+y=5 8. 15x+5y=2 9. 2m=-4n-4 10. 3m+5n=-3 11. 5x+y=0 12. 5x+2y=30 Thank you. (:
Tuesday, February 11, 2014 at 3:34pm by Melanie

Alegbra
3x - 2y = x - 6 3(x + 2y) = 3 Lets rearrange the second equation 3(x + 2y) = 3 x + 2y = 1 Now you have: 3x - 2y = x - 6 x + 2y = 1 You want to eliminate one variable and the easiest one to do so is y. So add the two equations together like so.. 3x - 2y + x + 2y = x - 6 + 1 ...
Thursday, April 3, 2008 at 1:29pm by Anonymous

algebra
A = (x+y)(x^2+6xy-4y^2), A = x^3+6x^2y-4xy^2+x^2y+6xy^2-4y^3, A = x^3 + 7x^2y + 2xy^2 - 4y^3 sq in.
Tuesday, July 19, 2011 at 11:17pm by Henry

Honors Geometry
1. If OB bisects <AOC, then <AOB + <BOC = <AOC. 6x + y + 5 = 2x + 2y solve for y and you get 4x + 5 = y 2. Also, since OB bisect <AOC, then m<AOB = m<BOC 6x = y + 5 Now, use what y equals from step one and subsitute it in so that: 6x = (4x + 5) + 5 ...
Wednesday, August 27, 2008 at 8:32pm by RCisME

Calculus
At what point(s) does the curve defined by x^2-6x=6-6y^2 have horizontal tangents? I took the derivative and solved for y'. I got y'=-1/6x/y+1/2y. hmm...what am I suppose to do?
Sunday, November 18, 2007 at 12:46am by Anonymous

Math
Eq1: 6X -Y = -4. Eq2: 2X + 2Y = 15. Solve for Y IN Eq1: 6X - Y = -4, -Y = -6X -4, Y = 6X + 4, Substitute 6x + 4 for Y IN Eq2: 2X + 2(6X + 4) = 15, 2X + 12X + 8 = 15, 14X= 15 - 8 = 7, X = 1/2. Substitute 1/2 for X IN Eq1: 6 * 1/2 - Y = -4, 3 - Y = -4, -Y = -7, Y = 7. Solution...
Sunday, February 13, 2011 at 3:42pm by Henry

algebra
or, you could recognize it as the sum of two cubes: 8y^12+z^6 = (2y^4)^3 + (z^2)^3 = (2y^4 + z^2)((2y^4)^2 - (2y^4)(z^2) + (z^2)^2) = (2y^4 + z^2)(4y^8 - 2y^4z^2 + z^4)
Tuesday, April 3, 2012 at 9:50pm by Steve

Calculus
let the width of the rectangle be 2x let the height of the rectangle by y then sides of triangle are 2x each and its height is √3 x area = (1/2((2x)(√3x) + 2xy = √3 x^2 + 2xy 200 = √3 x^2 + 2xy u = (200 - √3x^2)/(2x) Perimeter = P = 6x + 2y = 6x...
Sunday, February 16, 2014 at 11:24am by Reiny

Math
What is confusing. The problem says, "Kosuke got x correct" and then asks for "an expression for the number of questions he answered correctly " Ummm, x? For the score, just multiply the points by each type of question: If there were 25 questions, and he answered x and left y ...
Sunday, October 9, 2011 at 3:09am by Steve

Pre-algebra
1. (6x+1)(x^2 - 9x + 5). Multiply 6x by each term in the 2nd parenthesis. Then multiply 1 by each term in the 2nd parenthesis: 6x^3 -54x^2 + 30x + x^2 - 9x + 5, Combine like-terms: 6x^3 - 53x^2 +21x + 5. 2. Same procedure as #1. 3, (12y^3 + 8y^2 + 20y) / 2y. Divide each term ...
Saturday, December 10, 2011 at 3:30am by Henry

Math
Multiply first equation by 3 and the second by 2. 6x - 12y = 9 -6x + 10y = 2 Add the two equations. -2y = 11 You should be able to work it from here.
Friday, October 25, 2013 at 11:31am by PsyDAG

Algebra 2
F(x) = (6x-6)/4. g(x) = 1/x. (F/g)(x) = ((6x-6)/4) divided by 1/x, In vert and multiply: =((6x-6)/4)*x/1=x(6x-6)/4 = 6x(x-1)/4 = 3x(x-1)/2. F(x) = 6x-7. g(x) = 8x-3. (f-g)(x) = (6x-7)-(8x-3)= 6x-7-8x+3 = -2x-4 = -2(x+2). F(x) = 2-4x. g(x) = -6x+4. (f+g)(x)=(2-4x) + (-6x+4)=2-...
Monday, July 4, 2011 at 4:13pm by Henry

Math
-4y = x^2 + 6x +5 = x^2 + 6x + 9 - 9 + 5 = (x+3)^2 - 4 divide both sides by -4 y = (-1/4) (x+3)^2 + 1 try the second the same way. (looks like some kind of type, what does .. 2y+=-25 mean ?)
Monday, May 7, 2012 at 9:20pm by Reiny

algebra
Could you show me how to do these math systems and show your work so i understand how to do them. 1) 3x+2y+4z=19 2x-y+z=3 6x+7y-z=17 2) 5x+8y+6z=14 2x+5y+z=12 -3x+6y-12z=20 3) x^2+2xy-2y^2=-11 x^2+xy-2y^2=-9
Monday, April 8, 2013 at 1:38am by hal

math
Multiply the second equation by 2. 8x + 2y = 2 6x + 2y = -2 Subtract bottom equation from top. 2x = 4 Take it from there.
Tuesday, February 15, 2011 at 5:35pm by PsyDAG

algebra
I would divide first. notice the 6x^2 y^3 in the numerator, adn the denominatora 3x^3y^2. Reduce those: Now (2y)(9x^2y^3) is left which is easy to finish.
Thursday, June 9, 2011 at 1:38pm by bobpursley

math
by "work out" do you mean solve ? double the second and add to the first 5x-2y=4 6x+2y=18 11x = 22 x = 2 sub into original 2nd 3(2) + y = 9 y = 9-6 y = 3
Thursday, November 8, 2012 at 9:00pm by Reiny

Substution Soluton MATH
3. Assuming it was a typo and the 2nd equation was 4x+2y = 8, notice that both terms contain 2y as the y-term. The method I used is called "elimination". if we subtract the two equations, of course we can only add/subtract like terms, I get: 6x-4x = 2x 2y - 2y = 0 ----> Ahh...
Thursday, October 31, 2013 at 5:58am by Reiny

Algebra
Please don't capitalize your variables, it makes it hard to read them. (-x + y - 2(-2y)(x - y) )/( (x-y)^2 + (x+y)^2 ) = (-x + y + 4xy - 4y^2)/(x^2 - 2xy + y^2 + x^2 + 2xy + y^2) = ( x(4y - 1) - y(4x - 1) )/( 2(x^2 + y^2) ) = (4x-1)(x-y) / (2(x^2 + y^2) ) for the second, I don...
Sunday, May 5, 2013 at 6:55pm by Reiny

Algebra
(2/7x^2+7x^2y-4y^3)+(7/6x^2-2/5x^2y^2+5y^3)
Friday, September 16, 2011 at 11:38am by miranda

calculus
Differentiate both side with respect to x, implicitly. 6x -2y -2x dy/dx +2y dy/dx = 0 dy/dx*(x-y) = 3x -2y The tangent is vertical at x = y and horizontal at x = 2y/3. Solve the original equation for the corresponding x and y values. Horizontal tangent at: 3x^2+2x+x^2 = 24 x^2...
Monday, April 2, 2012 at 4:45am by drwls

derivative
I am finding the second derivative of (x^2)(y^3) = 1 2x(y^3) + (x^2)(y')(3y^2) = 0 And I simplified the answer to y'= (-2y)/(3x) so it would be... [(-2y')(3x)-(-2y)(3)]/(9x^2) and it's simplified to (-6xy'+6y)/(9x^2)= y'' I could plug in (-2y)/(3x) in place of y'. The correct ...
Monday, September 25, 2006 at 5:14pm by Jin

Algebra
-8.4x^3y^4 - 3.6x^2y^5 - 4.8x^4y^3 Factor out -1.2x^2y^3 to get -1.2x^2y^3(7xy + 3y^2 + 4x^2) Rearrange terms to get -1.2x^2y^3(4x^2 + 7xy + 3y^2) Now factor that to get -1.2x^2y^3(x+y)(4x+3y)
Monday, July 9, 2012 at 6:19am by Steve

intermeidate algebra
double the first -6x + 8y = 2 -1 times the 2nd 6x - 6y = 3 add them: 2y = 5 y = 5/2 sub into 1st: -3x + 4(5/2) = 1 -3x = -9 x=3
Friday, June 29, 2012 at 2:44pm by Reiny

math,algebra
I'm not understanding so the answer should read as follows and steps: 15x^4y^9 divided by 3x^2y^2 = 5x^2y^7 Can you check this one for me. Direction. Simplify each of the following expressions where possible. 6x^2y^3+9x^2y^3 divided by 3x^2y^2 My answer: 15x^4y^6 divided by 3x...
Monday, December 11, 2006 at 8:38pm by Jasmine20

algebra
how do I do this Solve for x. 3x – 2y = 6 Add 2y to both sides and then divide both sides by 3, to give you just "x' on the left side. so is this how i work it out 3x-2y=6 3x- 2y= 6 -2y 3x=6-2y x= 2 +2y No. Add 2y to both sides. 3x- 2y+2y= 6 +2y 3x=6+2y divide both sides by ...
Saturday, April 28, 2007 at 12:11am by help

Derivative
2xy' + 2y = 2y y' 2xy' - 2y y' = -2y y'(2x - 2y) = -2y y' = y/(y-x) y'' = ( (y-x)y' - y(y' - 1) )/(y-x)^2 = [ (y-x)(y/(y-x) ) - y( y/(y-x) - 1) ]/(y-x)^2 = [ y - y^2 /(y-x) + y ]/(y-x)^2 = [ y - y^2 /(y-x) + y ]/(y-x)^2 * (y-x)/y-x) =[ 2y(y-x) - y^2 ]/(y-x)^3 = (y^2 - 2x)/(y-x...
Thursday, March 21, 2013 at 10:03am by Reiny

Math Algebra
You need another 2y to get 4y 4y2 - 20y + 25 = (2y-5) (2y ) now that-5*2y = -10y so the product of -5 and the last term must be 25 so -5 so (2y-5)(2y-5)
Wednesday, April 1, 2009 at 1:18am by Damon

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