x^2+xy / 15y^3 * 35y^2 / 6x+6y * 12x / 10y^2

Guide :

x^2+xy / 15y^3 * 35y^2 / 6x+6y * 12x / 10y^2

(x^2+xy / 15y^3 ) (35y^2 / 6x + 6y) ( 12x / 10y^2)

Research, Knowledge and Information :


how to multiply rational expression of (x^2 + xy / 15y^3 ...


how to multiply rational expression All Activity; Q&A; Questions; Unanswered ... (x^2 + xy / 15y^3) (35y^2 / 6x+6y) (12x / 10y^2) 0 votes. how to multiply rational ...
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simplify the rational expression 2(x+4)(x-2)/3x^2-12x+12 ...


simplify the rational expression 2(x+4)(x-2)/3x^2-12x+12. 0 votes. ... (x^2 + xy / 15y^3) (35y^2 / 6x+6y) (12x / 10y^2) asked Nov 21, 2013 in ALGEBRA 2 by skylar ...
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4x 6y =12 and -6x 15y = 30? - Weknowtheanswer


4x 6y =12 and -6x 15y ... of the centre of the circle for the following x^2 + y^2 - 4x - 6y = ... 8y -12 ... 10y=10 -8-6y=6 ... X=0. Then for y 16x-10y=10. -16x-12y ...
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Find the equation of the circle passing through the center of ...


What is the equation of the circle concentric with the circle x^2+y^2+6x-10y+33=0 and ... x^2+y^2-4x-6y-9=0[/math ... The line x _ y = 0 cuts the circle x^2 +y^2 _ 2x ...
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Math 432 HW 2.5 Solutions


Math 432 HW 2.5 Solutions Assigned: 1-10, 12, 13, and 14. Selected for Grading: 1 (for five points), 6 (also for five), 9, 12 Solutions: 1. (2y3 + 2y2) dx + (3y2x ...
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SOLUTIONS - UCSD Mathematics


SOLUTIONS Problem 1. Find the ... xx = 12x 6y 24; B= f xy = 6x; C= f yy = 6: At the point (0;0) we have f ... 2(y+1) = y 1 =)y= 3: Now, g(x;y) = 18 =)x= 1: At ( 1; 3 ...
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Expression Factoring Calculator | Wyzant Resources


Expression Factoring Calculator. Expression: Example Expression ... Similarly, 2 * (x + 5) can also be entered as 2(x + 5); 2x * (5) can be entered as 2x(5).
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Suggested Questions And Answer :


5(2x + 1) = 12x – 1

5(2x + 1) = 12x - 1 10x + 5 = 12x - 1   Now you can subtract 5 from both sides of the equation: 10x + 5 - 5 = 12x - 1 - 5 10x = 12x - 6   Now you can subtract '12x' from both sides of the equation: 10x - 12x = 12x - 6 - 12x - 2x = - 6   Now
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what are the critical numbers of the equation: 3x^4-8x^3+6x^2

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Problem: 8x − 2 = y 5y − 12x = -3 Select the ordered pair from the choices below that is a solution to the following system of equations: 8x − 2 = y 5y − 12x = -3 1) 8x − 2 = y 2) 5y − 12x = -3 Equation 1 gives us the value of y, so substitute that into equation 2. 5y − 1
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8x-6y=14,12x-9y=18 Multiply each equation by the value that makes the coefficients of y equal.  This value is found by dividing the least common multiple of the coefficients of y by the current coefficient.  In this case, the least common multiple is 18. 3*(8x-6y=14)_2*(12x-9y=18) Multip
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What are the solutions to the equation:

y=(x)(x)(x+3)(x)(x-4)(x-12) of I wrote that correctly... man this is gonna be a pain... (x)(x) = x^2 (x^2)(x+3) = x^3 + 3x^2 (x^3 + 3x^2)(x) = (x^4 + 3x^3) (x^4 + 3x^3)(x-4) = (x^5 - 4x^4 + 3x^4 - 12x^3) put like terms together = (x^5 - x^4 - 12x^3) (x^5 - x^4 - 12x^3)(x-12) = re arrange to make eas
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(12x + 2)(3x − 6) What do you want to do? Expand or solve/ To expand: 12x(3x -6)  +2(3x - 6)  =  36x 2  -72x  + 6x  -12   ( i.e 36 x  squared) =   36x 2 - 66x  -12     To solve, it should be  (1
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Solve the second order ordinary differential equation y"-4y=12x y(0)=4 and y'(0)=1?

y"-4y=12x. Let y=p+q where p and q are functions of x, then y"=p"+q". So p"+q"-4p-4q=12x. If p"-4p=0 can be solved we only have to solve q"-4q=12x. p"-4p=0 has the characteristic equation (m-2)(m+2)=0 and we use the roots to suggest what p might be: p=Ae^2x+Be^-2x; p'=2Ae^2x-2Be^2x; p
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I appreciate for having shown your work. 3x-5(x+6)=3x-5x-30 12x-25_>3x-5x-30 12x-25_>-2x-30 12x+2x-25_>-2x+2x-30 14x-25_>30 14x-25+25_>-30+25 14x_>-5 14/14x_>=-5/14 x_>-5/14 The solution in set builder notation is {xIx_>-5/14}  
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How do I solve this?

How do I solve this? 7x-6y=22 12x-8y=1 1) 7x - 6y = 22 2) 12x - 8y = 1 Solve equation 1 for y in terms of x. 7x - 6y = 22 -6y = -7x + 22 y = 7/6 x - 22/6 Substitute that into equation 2, in place of y. 12x - 8y = 1 12x - 8(7/6 x - 22/6) = 1 12x - 4(7/3 x) + 4(22/3) = 1 12x - 28/3 x + 88/3 = 1 3
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Solve the second order ordinary differential equation

y"-4y=12x. Let y=p+q where p and q are functions of x, then y"=p"+q". So p"+q"-4p-4q=12x. If p"-4p=0 can be solved we only have to solve q"-4q=12x. p"-4p=0 has the characteristic equation (m-2)(m+2)=0 and we use the roots to suggest what p might be: p=Ae^2x+Be^-2x; p'=2Ae^2x-2Be^2x; p
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