How do I put y = -2x^2 + 5x - 10 in y = a(x-h)^2 + k form?
Q. How do I put y = -2x^2 + 5x - 10 in y = a(x-h)^2 + k form?
Asked by Jake - Thu Feb 7 18:41:25 2008 - - 1 Answers - 0 Comments
A. 4 hours...y = -2 (x^2 -[5/2] x) - 10 = -2 ( x^2 -2.5x + [ -2.5/2]^2) + 2 [ -2.5/2]^2 - 10 = -2 { x-5/4} ^ 2 - 55/8...note (x+h)^2 = x^2 +2hx + h^2...thus take half the coef of the x term and square it...of course what you add you then must subtract.
Answered by ted s - Thu Feb 7 23:15:19 2008
Q. How do I put y = -2x^2 + 5x - 10 in y = a(x-h)^2 + k form?
Asked by Jake - Thu Feb 7 18:41:25 2008 - - 1 Answers - 0 Comments
A. 4 hours...y = -2 (x^2 -[5/2] x) - 10 = -2 ( x^2 -2.5x + [ -2.5/2]^2) + 2 [ -2.5/2]^2 - 10 = -2 { x-5/4} ^ 2 - 55/8...note (x+h)^2 = x^2 +2hx + h^2...thus take half the coef of the x term and square it...of course what you add you then must subtract.
Answered by ted s - Thu Feb 7 23:15:19 2008
derivative of (5x^2+2x+2)/(sqrt(x))?
Q. (sqrt(x)(10x+2)-(5x^2+2x+ 2)(.5x^-.5))/x is not working
Asked by kyle - Thu Oct 9 16:41:11 2008 - - 2 Answers - 0 Comments
A. Divide through by the sqrt(x) = x^(1/2). (5x^2 + 2x + 2)/sqrt(x) = 5x^(3/2) + 2x^(1/2) + 2x^(-1/2) Now just apply the power rule to each part. Can you take it from there? P.S. What you were trying with the quotient rule was correct and if tidied up would give the same answer.
Answered by mathsmanretired - Thu Oct 9 16:51:06 2008
Q. (sqrt(x)(10x+2)-(5x^2+2x+ 2)(.5x^-.5))/x is not working
Asked by kyle - Thu Oct 9 16:41:11 2008 - - 2 Answers - 0 Comments
A. Divide through by the sqrt(x) = x^(1/2). (5x^2 + 2x + 2)/sqrt(x) = 5x^(3/2) + 2x^(1/2) + 2x^(-1/2) Now just apply the power rule to each part. Can you take it from there? P.S. What you were trying with the quotient rule was correct and if tidied up would give the same answer.
Answered by mathsmanretired - Thu Oct 9 16:51:06 2008
2x^2 + 3x +1 / x^2 + 1x - 6..and then that whole thing AGAIN DIVIDED by x^2 + 4x + 3 / 2x^2 - 5x + 2 ?
Q. the final answer for the numerator i got is (4x^2-1)...but the denominator is not (2x^2+5x-3) Please anyone help
Asked by Tomo M - Fri Sep 5 10:36:14 2008 - - 2 Answers - 0 Comments
A. (4x^2-1)/(x^2+6x+9)
Answered by butterflyfairy28 - Fri Sep 5 11:04:05 2008
Q. the final answer for the numerator i got is (4x^2-1)...but the denominator is not (2x^2+5x-3) Please anyone help
Asked by Tomo M - Fri Sep 5 10:36:14 2008 - - 2 Answers - 0 Comments
A. (4x^2-1)/(x^2+6x+9)
Answered by butterflyfairy28 - Fri Sep 5 11:04:05 2008
How do you factor: 2x^2 + 5x + 3?
Q. I dont understand factoring very well. How do you do these problems? 1. 2x^2 + 5x + 3 2. 3x^2 - 10x + 3 Can someone show me the steps on how to do this? I have a math final tomorrow.
Asked by kairi - Tue May 19 00:35:50 2009 - - 2 Answers - 0 Comments
A. (x+1)(2x+3) (x-3)(3x-1)
Answered by Rich - Tue May 19 00:40:24 2009
Q. I dont understand factoring very well. How do you do these problems? 1. 2x^2 + 5x + 3 2. 3x^2 - 10x + 3 Can someone show me the steps on how to do this? I have a math final tomorrow.
Asked by kairi - Tue May 19 00:35:50 2009 - - 2 Answers - 0 Comments
A. (x+1)(2x+3) (x-3)(3x-1)
Answered by Rich - Tue May 19 00:40:24 2009
I'm stumped. How do you simplify 2x^2+5x+2/2x^2+x-6 6x+3/2x-3 ?
Q. Those are both fractions, FYI. If you can explain it/figure it out, I'd appreciate it.
Asked by Hayley M - Sun Aug 24 05:28:48 2008 - - 1 Answers - 0 Comments
A. Well after alot of writing, its: ((4x^4)+(12x^3)-(6x^2)+x+ 2) divided by (2x^2).
Answered by billythekid - Sun Aug 24 05:44:48 2008
Q. Those are both fractions, FYI. If you can explain it/figure it out, I'd appreciate it.
Asked by Hayley M - Sun Aug 24 05:28:48 2008 - - 1 Answers - 0 Comments
A. Well after alot of writing, its: ((4x^4)+(12x^3)-(6x^2)+x+ 2) divided by (2x^2).
Answered by billythekid - Sun Aug 24 05:44:48 2008
Use the ploynomial below to answer the following = X^3 + 2x^2 -5x -6?
Q. Use the polynomial below to answer the following. 1)X^3 + 2x^2 -5x -6 a) List the real zeros for x b) use the reali zeros from d) express F(x) in factored form. Please help!!!
Asked by sunateastblvd - Thu May 8 03:22:02 2008 - - 1 Answers - 0 Comments
A. a) {3,-1,2} b) (x-3)(x+1)(x-2)
Answered by phoenix - Thu May 8 03:28:45 2008
Q. Use the polynomial below to answer the following. 1)X^3 + 2x^2 -5x -6 a) List the real zeros for x b) use the reali zeros from d) express F(x) in factored form. Please help!!!
Asked by sunateastblvd - Thu May 8 03:22:02 2008 - - 1 Answers - 0 Comments
A. a) {3,-1,2} b) (x-3)(x+1)(x-2)
Answered by phoenix - Thu May 8 03:28:45 2008
How do you factorise fully: x^3 + 2x^2 - 5X - 6?
Q. Hi, im doing further maths AS and i need some help doing the above question. I think sayin let f(x)=x^3 + 2x^2 - 5X - 6 would be the right track but I dont know what to do after that. Any help would before tomorow would be much appreciated!
Asked by omar f - Tue Oct 6 15:04:18 2009 - - 2 Answers - 0 Comments
Q. Hi, im doing further maths AS and i need some help doing the above question. I think sayin let f(x)=x^3 + 2x^2 - 5X - 6 would be the right track but I dont know what to do after that. Any help would before tomorow would be much appreciated!
Asked by omar f - Tue Oct 6 15:04:18 2009 - - 2 Answers - 0 Comments
3x^2 (2x^2 - 5x + 3)=?
Q. (a+2) (a^2 - 3a + 7)= (xy + 4)(xy-3)= (10-b)(10+b)= 6a^2 - 15a= y^2 - 9= x^2 + 7x - 18= xy + 4y - 2x -8= x^2 - 12x + 36= x^3 + 125= 8x^5 - 98x^3= a^2 - 15a + 56= Yes Yes I know! You don't want to do my homework, but I just need HELP getting in the right direction. I get so confused in all this, all the stuff looks the same and I cannot get a correct answer. Or at least I don't think it's right. Help me out please!
Asked by Angelica - Wed Sep 26 14:37:39 2007 - - 1 Answers - 0 Comments
A. Ok, the first two are all very similar in that you are simply distributing the front expression into the parentheses. So, 3x^2*(2x^2 - 5x + 3) = 3x^2*2x^2 - 3x^2*5x + 3x^2*3 = 6x^4 - 15x^3 + 9x^2 For the next two, again, they are similar. The expressions are both binomials, so you need to FOIL out it out. Remember, FOIL stands for First, Outside, Inside, Last. Meaning, you must multiply the First terms, multiply the Outside terms, multiply the Inside terms, and multiply the Last terms. So for your third problem you get: (xy + 4)(xy - 3) = (xy)*(xy) + (xy)*(-3) + 4*(xy) + 4*(-3) = x^2*y^2 - 3xy + 4xy - 12 = x^2*y^2 + 1xy - 12, or x^2*y^2 + xy - 12 The remaining problems are all factoring problems. I'll pick out a few different cases… [cont.]
Answered by Lee - Wed Sep 26 17:00:35 2007
Q. (a+2) (a^2 - 3a + 7)= (xy + 4)(xy-3)= (10-b)(10+b)= 6a^2 - 15a= y^2 - 9= x^2 + 7x - 18= xy + 4y - 2x -8= x^2 - 12x + 36= x^3 + 125= 8x^5 - 98x^3= a^2 - 15a + 56= Yes Yes I know! You don't want to do my homework, but I just need HELP getting in the right direction. I get so confused in all this, all the stuff looks the same and I cannot get a correct answer. Or at least I don't think it's right. Help me out please!
Asked by Angelica - Wed Sep 26 14:37:39 2007 - - 1 Answers - 0 Comments
A. Ok, the first two are all very similar in that you are simply distributing the front expression into the parentheses. So, 3x^2*(2x^2 - 5x + 3) = 3x^2*2x^2 - 3x^2*5x + 3x^2*3 = 6x^4 - 15x^3 + 9x^2 For the next two, again, they are similar. The expressions are both binomials, so you need to FOIL out it out. Remember, FOIL stands for First, Outside, Inside, Last. Meaning, you must multiply the First terms, multiply the Outside terms, multiply the Inside terms, and multiply the Last terms. So for your third problem you get: (xy + 4)(xy - 3) = (xy)*(xy) + (xy)*(-3) + 4*(xy) + 4*(-3) = x^2*y^2 - 3xy + 4xy - 12 = x^2*y^2 + 1xy - 12, or x^2*y^2 + xy - 12 The remaining problems are all factoring problems. I'll pick out a few different cases… [cont.]
Answered by Lee - Wed Sep 26 17:00:35 2007
How would you factor the equation of 2x^2 + 5x - 3?
Q. Self explanatory. I'm not exactly sure what numbers fit this equation. I know it goes along the lines of (2x +- ?)(x +- ?), but that's as far as I know. How would you get it?
Asked by Mark R - Mon Oct 19 22:33:38 2009 - - 4 Answers - 0 Comments
A. if you have a function ax^2 + bx + c = 0 consider (2x+a)(x+b) expand this gives 2x^2 + 2bx + ax +ab =0 so looking at the coefficients we need: ab = -3 2b + a = 5 so -3 * 1 or 3 * -1 gives -3 so we have 4 cases: a = -1 b = 3 a = 1 b = -3 a = 3 b = -1 a = -3 b = 1 now consider 2b + a = 5 2(1) + (-3) = -1 2(-1) + (3) = 1 2(3) + (-1) = 5 this is what we want. So we have b = 3 and a = -1 which gives the factorization (2x-1)(x+3) or you could use the quadratic formula (-b +- sqrt(b^2 - 4(a*c))) / (2(a)) where your formula is ax^2 + bx + c = 0 gives: (-5 +- sqrt(25-4(-6)))/(2(2)) = (-5 +- sqrt(49))/4 = (-5 + 7)/4 or (-5 -7)/4 = 1/2 or -3 => (x+(1/2))(x-3) expand this (then multiply the whole equation by 2) and you find it… [cont.]
Answered by crazylady - Mon Oct 19 22:48:57 2009
Q. Self explanatory. I'm not exactly sure what numbers fit this equation. I know it goes along the lines of (2x +- ?)(x +- ?), but that's as far as I know. How would you get it?
Asked by Mark R - Mon Oct 19 22:33:38 2009 - - 4 Answers - 0 Comments
A. if you have a function ax^2 + bx + c = 0 consider (2x+a)(x+b) expand this gives 2x^2 + 2bx + ax +ab =0 so looking at the coefficients we need: ab = -3 2b + a = 5 so -3 * 1 or 3 * -1 gives -3 so we have 4 cases: a = -1 b = 3 a = 1 b = -3 a = 3 b = -1 a = -3 b = 1 now consider 2b + a = 5 2(1) + (-3) = -1 2(-1) + (3) = 1 2(3) + (-1) = 5 this is what we want. So we have b = 3 and a = -1 which gives the factorization (2x-1)(x+3) or you could use the quadratic formula (-b +- sqrt(b^2 - 4(a*c))) / (2(a)) where your formula is ax^2 + bx + c = 0 gives: (-5 +- sqrt(25-4(-6)))/(2(2)) = (-5 +- sqrt(49))/4 = (-5 + 7)/4 or (-5 -7)/4 = 1/2 or -3 => (x+(1/2))(x-3) expand this (then multiply the whole equation by 2) and you find it… [cont.]
Answered by crazylady - Mon Oct 19 22:48:57 2009
What is the area enclosed by the set of curves: y=3x^2-9x+17, y=2x^2+5x-16?
Q. please help! I'm home from school, forgot my graphing calculator and the answers I am coming up with seem unlikely!
Asked by thelonemonkee - Tue Nov 7 20:08:49 2006 - - 2 Answers - 0 Comments
A. To find the area enclosed by two curves, you first set the two curves equal to each other so that you can determine the two endpoints of your curve: 3x - 9x + 17 = 2x + 5x - 16, implies: x - 14x + 33 = 0, or (x-3)(x-11) = 0 So the endpoints of your curve occur at x=3 and x=11 So now you want to see if these curves are both in the upper quadrant between x=3 and x=11, or both in the lower quadrant, or one in the upper and one in the lower. By creating a little Excel formula and plugging in all the integers from 3 to 11, I see that both curves are in the first quadrant and the second one is strictly above the first one for all points between 3 and 11 (except where they are exactly equal at 3 and 11). Therefore to find the area… [cont.]
Answered by I AUG - Tue Nov 7 23:33:49 2006
Q. please help! I'm home from school, forgot my graphing calculator and the answers I am coming up with seem unlikely!
Asked by thelonemonkee - Tue Nov 7 20:08:49 2006 - - 2 Answers - 0 Comments
A. To find the area enclosed by two curves, you first set the two curves equal to each other so that you can determine the two endpoints of your curve: 3x - 9x + 17 = 2x + 5x - 16, implies: x - 14x + 33 = 0, or (x-3)(x-11) = 0 So the endpoints of your curve occur at x=3 and x=11 So now you want to see if these curves are both in the upper quadrant between x=3 and x=11, or both in the lower quadrant, or one in the upper and one in the lower. By creating a little Excel formula and plugging in all the integers from 3 to 11, I see that both curves are in the first quadrant and the second one is strictly above the first one for all points between 3 and 11 (except where they are exactly equal at 3 and 11). Therefore to find the area… [cont.]
Answered by I AUG - Tue Nov 7 23:33:49 2006
How would you factor 2x^2 + 5x + 3?
Q. I know how to do this one: x^2 + 6x + 8 = (x +2)(x +4), but the 2x^2 is throwing me off!
Asked by GothMimiKitty - Tue Feb 13 21:58:54 2007 - - 7 Answers - 0 Comments
A. 1. You need to find two numbers whose product is 2 * 3 (the first and last coefficients) and whose sum is 5 (the second coefficient) The numbers are 2 and 3. 2. Split the 5x into two parts using the numbers you found. 2x^2 + 5x + 3 = 2x^2 + 2x + 3x + 3 3. Now, by grouping two terms at a time, factor out the largest factor: 2x^2 + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) 4. Then, factor out the common factor (x+1) (x+1)(2x +3) If you do not understand step 4, think of it this way: let y = (x=1) 2xy + 3y = y(2x+3) Now, replace y with x+1 (x+1)(2x+3)
Answered by DSS - Tue Feb 13 22:05:44 2007
Q. I know how to do this one: x^2 + 6x + 8 = (x +2)(x +4), but the 2x^2 is throwing me off!
Asked by GothMimiKitty - Tue Feb 13 21:58:54 2007 - - 7 Answers - 0 Comments
A. 1. You need to find two numbers whose product is 2 * 3 (the first and last coefficients) and whose sum is 5 (the second coefficient) The numbers are 2 and 3. 2. Split the 5x into two parts using the numbers you found. 2x^2 + 5x + 3 = 2x^2 + 2x + 3x + 3 3. Now, by grouping two terms at a time, factor out the largest factor: 2x^2 + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) 4. Then, factor out the common factor (x+1) (x+1)(2x +3) If you do not understand step 4, think of it this way: let y = (x=1) 2xy + 3y = y(2x+3)
Answered by DSS - Tue Feb 13 22:05:44 2007
Subtract (x^2 9x + 8) from ( -2x^2 5x + 3).?
Q. Subtract (x^2 9x + 8) from ( -2x^2 5x + 3).
Asked by Tyler - Fri Apr 24 16:49:09 2009 - - 1 Answers - 0 Comments
A. To do this, you just subtract like terms to get -3x^2 + 4x -5
Answered by Mather - Fri Apr 24 18:14:39 2009
Q. Subtract (x^2 9x + 8) from ( -2x^2 5x + 3).
Asked by Tyler - Fri Apr 24 16:49:09 2009 - - 1 Answers - 0 Comments
A. To do this, you just subtract like terms to get -3x^2 + 4x -5
Answered by Mather - Fri Apr 24 18:14:39 2009
For the function f defined by f(x)= 2x^2+5x+7 Find the following values?
Q. For the function f defined by f(x)= 2x^2+5x+7 Find the following values.. a) f(-7) b) f(-x) c) -f(x) d) f(x+h)
Asked by Timothy - Wed Sep 16 16:32:24 2009 - - 2 Answers - 0 Comments
A. a) f(-7)=2(-7)^2+5(-7)+7 =2(49)+(-35)+7 =98-28 =70 b) f(-x)=2(-x)^2+5(-x)+7 =2x^2-5x+7 c) -f(x)= -2x^2-5x-7 Not sure about that one though, don't take my word exactly for it d) f(x+h)=2(x+h)^2+5(x+h)+7 =2(x^2+2xh+h^2)+5x+5h+7 =2x^2+h^2+2xh+5h+5x+7 Make sure to double check these, I am by no means perfect :)
Answered by unknown - Wed Sep 16 16:48:31 2009
Q. For the function f defined by f(x)= 2x^2+5x+7 Find the following values.. a) f(-7) b) f(-x) c) -f(x) d) f(x+h)
Asked by Timothy - Wed Sep 16 16:32:24 2009 - - 2 Answers - 0 Comments
A. a) f(-7)=2(-7)^2+5(-7)+7 =2(49)+(-35)+7 =98-28 =70 b) f(-x)=2(-x)^2+5(-x)+7 =2x^2-5x+7 c) -f(x)= -2x^2-5x-7 Not sure about that one though, don't take my word exactly for it d) f(x+h)=2(x+h)^2+5(x+h)+7 =2(x^2+2xh+h^2)+5x+5h+7 =2x^2+h^2+2xh+5h+5x+7 Make sure to double check these, I am by no means perfect :)
Answered by unknown - Wed Sep 16 16:48:31 2009
Use the quadratic formula to solve the equation 2x^2-5x=-7?
Q. The solution set is {?} Must be an exact answer, using radicals as needed, and complex numbers must be in terms of i. Can use integers and fractions for any numbers in the expression.
Asked by Plz2HelpMe!KKTHXU - Sat Apr 4 15:23:30 2009 - - 3 Answers - 0 Comments
A. Put it in standard quadratic form (ax^2 + bx + c = 0) by adding 7 to both sides: 2x^2 - 5x + 7 = 0 a = 2; b = -5; c = 7 Use the quadratic formula: (-b +/- sqrt(b^2 - 4ac))/2a (-(-5) +/- sqrt((-5)^2 - 4(2)(7)))/2(2) (5 +/- sqrt(-31))/4 (5 +/- sqrt(-1)(sqrt31))/4 sqrt-1 = i, so: (5 +/- isqrt31)/4 The solution set is {(5 + isqrt31)/4, (5 - isqrt31)/4}<===ANSWER Since this is an imaginary number, the solution set could also be {no real number}<===Answer bl
Answered by GunsUpTech - Sat Apr 4 15:31:54 2009
Q. The solution set is {?} Must be an exact answer, using radicals as needed, and complex numbers must be in terms of i. Can use integers and fractions for any numbers in the expression.
Asked by Plz2HelpMe!KKTHXU - Sat Apr 4 15:23:30 2009 - - 3 Answers - 0 Comments
A. Put it in standard quadratic form (ax^2 + bx + c = 0) by adding 7 to both sides: 2x^2 - 5x + 7 = 0 a = 2; b = -5; c = 7 Use the quadratic formula: (-b +/- sqrt(b^2 - 4ac))/2a (-(-5) +/- sqrt((-5)^2 - 4(2)(7)))/2(2) (5 +/- sqrt(-31))/4 (5 +/- sqrt(-1)(sqrt31))/4 sqrt-1 = i, so: (5 +/- isqrt31)/4 The solution set is {(5 + isqrt31)/4, (5 - isqrt31)/4}<===ANSWER Since this is an imaginary number, the solution set could also be {no real number}<===Answer bl
Answered by GunsUpTech - Sat Apr 4 15:31:54 2009
Explain how the discriminant can be used to show that the line 4x + 4y = 4 and the parabola y = 2x^2 - 5x + 6?
Q. ...do not intersect. The discriminants are 16 for the former and -23 for the latter (discriminant is b^2 - 4ac). My friend said that since the parabola only had an imaginary solution, that it did not cross the x axis, but that doesn't mean they don't intersect.
Asked by John B - Tue Sep 4 18:13:52 2007 - - 1 Answers - 0 Comments
A. You can learn math well, if you may follow the definitions and concepts well. The definition of "discriminant" is b^2 - 4ac in a quadratic equation ax^2 + bx + c = 0. Your argument is not valid since neither the line 4x + 4y = 4, nor the parabola y = 2x^2 - 5x + 6, is a quadratic equation. The parabola can be plotted as a curve in the x-y plan, such that every point is REAL. The line 4x + 4y = 4 can also be plotted in the x-y plan, such that every point is REAL. If they have a cross-section, the cross-section point/s must be REAL, since the cross-section point/s must be on both curve and the line. Since the cross-section must satisfy both equations, we may put: y = 1 - x (the line) into y = 2x^2 - 5x + 6 (the parabola), and get: 1 - x… [cont.]
Answered by Hahaha - Wed Sep 5 15:28:04 2007
Q. ...do not intersect. The discriminants are 16 for the former and -23 for the latter (discriminant is b^2 - 4ac). My friend said that since the parabola only had an imaginary solution, that it did not cross the x axis, but that doesn't mean they don't intersect.
Asked by John B - Tue Sep 4 18:13:52 2007 - - 1 Answers - 0 Comments
A. You can learn math well, if you may follow the definitions and concepts well. The definition of "discriminant" is b^2 - 4ac in a quadratic equation ax^2 + bx + c = 0. Your argument is not valid since neither the line 4x + 4y = 4, nor the parabola y = 2x^2 - 5x + 6, is a quadratic equation. The parabola can be plotted as a curve in the x-y plan, such that every point is REAL. The line 4x + 4y = 4 can also be plotted in the x-y plan, such that every point is REAL. If they have a cross-section, the cross-section point/s must be REAL, since the cross-section point/s must be on both curve and the line. Since the cross-section must satisfy both equations, we may put: y = 1 - x (the line) into y = 2x^2 - 5x + 6 (the parabola), and get: 1 - x… [cont.]
Answered by Hahaha - Wed Sep 5 15:28:04 2007
Evaluate the following limits lim x->oo 4x^-2/2x^2+5x-1?
Q. Evaluate the following limits lim x->oo 4x^-2/2x^2+5x-1?
Asked by HK - Thu Jun 12 00:52:33 2008 - - 2 Answers - 0 Comments
A. lim x->infinity (4x^(-2))/(2x^2+5x-1) = lim x->infinity 1/[(4x^(2))(2x^2+5x-1)] = 1/infinity = 0
Answered by MaX - Thu Jun 12 01:04:43 2008
Q. Evaluate the following limits lim x->oo 4x^-2/2x^2+5x-1?
Asked by HK - Thu Jun 12 00:52:33 2008 - - 2 Answers - 0 Comments
A. lim x->infinity (4x^(-2))/(2x^2+5x-1) = lim x->infinity 1/[(4x^(2))(2x^2+5x-1)] = 1/infinity = 0
Answered by MaX - Thu Jun 12 01:04:43 2008
What are the values for a, b, and c in the function y=-2x^2=+5x-4?
Q. What are the values for a, b, and c in the function y=-2x^2=+5x-4?
Asked by Trinibby - Thu Jul 30 22:32:09 2009 - - 6 Answers - 0 Comments
A. a is always the coefficient infront of x^2 b is always the coefficient infront of x c is always the number that doesnt have a coefficient so in your function: a = -2 b = 5 c = -4 also be sure to remember that if you were given the function: y = -5x^2 + 3x then the values for a, b and c would be: a = -5 b = 3 c = 0
Answered by STB - Thu Jul 30 22:38:57 2009
Q. What are the values for a, b, and c in the function y=-2x^2=+5x-4?
Asked by Trinibby - Thu Jul 30 22:32:09 2009 - - 6 Answers - 0 Comments
A. a is always the coefficient infront of x^2 b is always the coefficient infront of x c is always the number that doesnt have a coefficient so in your function: a = -2 b = 5 c = -4 also be sure to remember that if you were given the function: y = -5x^2 + 3x then the values for a, b and c would be: a = -5 b = 3 c = 0
Answered by STB - Thu Jul 30 22:38:57 2009
2x^2-5x+3=0 Solve by factoring, give x1 and or x2?
Q. 2x^2-5x+3=0 Solve by factoring, give x1 and or x2
Asked by Colton W - Sat Mar 28 18:16:37 2009 - - 5 Answers - 0 Comments
A. We have a negative middle term and a positive last term, so we're looking for something in the form of (?x - ?)(?x - ?) The only factors of 3 are 1 and 3, so this makes it (?x - 1)(?x - 3) The only factors of 2 are 1*2, so this means we either have (2x - 1)(x - 3) or (x - 1)(2x - 3) The only one of these that gives us -5x in the middle is (x - 1)(2x - 3). So if (x - 1)(2x - 3) = 0, then that means x-1=0 or 2x-3=0, so x is 1 or 3/2
Answered by Geezah - Sat Mar 28 18:23:01 2009
Q. 2x^2-5x+3=0 Solve by factoring, give x1 and or x2
Asked by Colton W - Sat Mar 28 18:16:37 2009 - - 5 Answers - 0 Comments
A. We have a negative middle term and a positive last term, so we're looking for something in the form of (?x - ?)(?x - ?) The only factors of 3 are 1 and 3, so this makes it (?x - 1)(?x - 3) The only factors of 2 are 1*2, so this means we either have (2x - 1)(x - 3) or (x - 1)(2x - 3) The only one of these that gives us -5x in the middle is (x - 1)(2x - 3). So if (x - 1)(2x - 3) = 0, then that means x-1=0 or 2x-3=0, so x is 1 or 3/2
Answered by Geezah - Sat Mar 28 18:23:01 2009
x^3+2x^2-3x-4..?? and x^4-5x^2+2x+2..??
Q. Can someone convert these 2 function into the factored form please? Thanks.
Asked by Mohika P - Thu Feb 28 16:22:44 2008 - - 4 Answers - 0 Comments
A. x^3+2x^2-3x-4 By trial and error, -1 is a solution, so (x+1) is a factor. Divide x^3+2x^2-3x-4 by (x+1) The quotient is (x^2+x-4), so the factorization is (x^2+x-4)(x+1). You cannot factor (x^2+x-4) unless you use a quadratic equation to find the factors which are fractional numbers. x^4-5x^2+2x+2 By trial and error 1 is a solution, so (x-1) is a factor. Divide x^4-5x^2+2x+2 by (x-1) we get (x^3+x^2-4x-2)(x-1) (x^3+x^2-4x-2) cannot be factored further.
Answered by cidyah - Thu Feb 28 16:49:25 2008
Q. Can someone convert these 2 function into the factored form please? Thanks.
Asked by Mohika P - Thu Feb 28 16:22:44 2008 - - 4 Answers - 0 Comments
A. x^3+2x^2-3x-4 By trial and error, -1 is a solution, so (x+1) is a factor. Divide x^3+2x^2-3x-4 by (x+1) The quotient is (x^2+x-4), so the factorization is (x^2+x-4)(x+1). You cannot factor (x^2+x-4) unless you use a quadratic equation to find the factors which are fractional numbers. x^4-5x^2+2x+2 By trial and error 1 is a solution, so (x-1) is a factor. Divide x^4-5x^2+2x+2 by (x-1) we get (x^3+x^2-4x-2)(x-1) (x^3+x^2-4x-2) cannot be factored further.
Answered by cidyah - Thu Feb 28 16:49:25 2008
How do I factor this polynomial completely: 2x^2-5x-3?
Q. How do I factor this polynomial completely: 2x^2-5x-3?
Asked by Ree R - Wed Jan 14 02:38:47 2009 - - 4 Answers - 0 Comments
A. 2x^2-5x-3 first set up brackets and factors for 2x^2 and 3 2x^2 can only be 2x times x and 3 can only be 3 times 1, but a negative 3 can be -3 times 1 or negative 1 times 3 so asses the equation and see how to proceed notice the middle term is -5 so your factors have to give you a -5 for your middle term remember the identity for factoring algebraic equations is (a+b)(a+b) multiply a in the first set of brackets by each of the numbers in the second set of brackets and repeat for the b. so you get a^2 +ba +ba = b^2=a^2 + 2ba +b^2 (a-b)(a+b)=a^2-ba+ba-b^2= a^2-b^2 (2x+1)(x-3)=0
Answered by patman6@sbcglobal.net - Wed Jan 14 02:52:58 2009
Q. How do I factor this polynomial completely: 2x^2-5x-3?
Asked by Ree R - Wed Jan 14 02:38:47 2009 - - 4 Answers - 0 Comments
A. 2x^2-5x-3 first set up brackets and factors for 2x^2 and 3 2x^2 can only be 2x times x and 3 can only be 3 times 1, but a negative 3 can be -3 times 1 or negative 1 times 3 so asses the equation and see how to proceed notice the middle term is -5 so your factors have to give you a -5 for your middle term remember the identity for factoring algebraic equations is (a+b)(a+b) multiply a in the first set of brackets by each of the numbers in the second set of brackets and repeat for the b. so you get a^2 +ba +ba = b^2=a^2 + 2ba +b^2 (a-b)(a+b)=a^2-ba+ba-b^2= a^2-b^2 (2x+1)(x-3)=0
Answered by patman6@sbcglobal.net - Wed Jan 14 02:52:58 2009
From Yahoo Answer Search: '2x^2 5x 2'
Mon Nov 9 08:26:38 2009 [ refresh local cache ]
