Friday

April 18, 2014

April 18, 2014

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*

**math-please I really need help!!!!!**

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*

**Math- Adv. Alg. w/ Trig.**

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.*

** math please help**

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*

** math please help**

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*

** math please help**

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*

**10 grade algebra**

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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