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some algebra help (radicals)
I hope I am writing this down right.. I am trying to do some practice questions to learn 10^5 (sqrt)2y  4^5 (sqrt)2y I am trying to figure out how to solve this They gave us some answers to choose from, but I am clueless on how
asked by Cassie on May 6, 2007 
Algebra
Solve for s: h=(square root of 3)times s/2 and solve for h V= (pi)r squared h / 3 Solve for s: h=(square root of 3)times s/2 Multiply both sides by 2. 2h = (sqrt 3)*s*2/2 which cancels the 2 on the right. 2h = (sqrt 3)*s Now
asked by Jamie on August 17, 2006 
Algebra 2: Radicals URGENT!!
Could some kind, saintly soul help me solve this problem? Simplify: 8w sqrt(48w^5)  x^2 sqrt(3xw^2) . . =8w(√16)(√3)(√w^4)(√w)  x^2(√3)(√x)(√w^2) =32w^3(√3w)  wx^2(√3x) not much of a "simplification" really 8w
asked by Maria on April 27, 2007 
algebra
can someone tell me if i did the problem correct (solve by completing the square) 4x^2+2x3=0 x^2+(1/2)=3/4 x^2+(1/2)+(1/4)^2=3/4+(1/4)^2 (x+1/4)^2=13/16 x+1/4=+sqrt 13/16 x+1/4=+sqrt 13/4 x=1/4+sqrt 13/4 x=1+sqrt 13/4
asked by kate on November 18, 2007 
Calculus
Please look at my work below: Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(1)(6)))/2(1) r=(16 +/ Sqrt(8)) r=8 +/ Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2,
asked by COFFEE on July 10, 2007 
Math/Calculus
Solve the initialvalue problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else? Thanks. y''+4y'+6y=0, y(0)=2, y'(0)=4 r^2+4r+6=0, r=(4 +/ sqrt(164(1)(6))/2 r=2 +/ sqrt(2)*i , alpha
asked by COFFEE on July 12, 2007 
Geometry
Okay, this is a problem that I've tried over ten times before getting a fresh problem and trying that one too. I've never gotten a correct answer and I can't figure out why it's not right. Here's the problem, "Find the distance
asked by Aubry on September 21, 2006 
math,help,algebra I
I need help can someone help me get unstuck and let me know if i am correct.thank you. solve by completing the square. 4x^2+2x3=0 this is what i am doing: i used the quadratic equation. so where i am is in this step: x = (2 (+)
asked by jasmine20 on April 14, 2007 
math
solve 2x^2+3x+8=0 and express the solutions in a+bi form. Let's use the quadratic formula to solve for x and express those solutions in a+bi form. x = [b + or  sqrt(b^2  4ac)]/2a Note: sqrt = square root. a = 2, b = 3, and c =
asked by Jimmy on January 10, 2007 
Calculus  Second Order Differential Equations
Posted by COFFEE on Monday, July 9, 2007 at 9:10pm. download mp3 free instrumental remix Solve the initialvalue problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/ Sqrt(4^24(1)(6)))/2(1) r=(16 +/ Sqrt(8))
asked by COFFEE on July 10, 2007