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 ...

### 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 ...

### 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 ...

### 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 ...

### 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 ...

### 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 ...

### 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).

## 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 you can divide both sides of the equation by ' -2 ' to find the value of 'x' on it's own: (- 2x)/(-2) = (- 6)/(-2) x = 6/2 = 3   So the final answer is x = 3 .

### what are the critical numbers of the equation: 3x^4-8x^3+6x^2

f(x) = 3x^4 - 8x^3 + 6x^2         Find the derivative of the function f'(x) = 12x^3 - 24x^2 +12x       The critical values are the f' set = 0   12x^3 -24x^2 + 12x  = 0               12x(x^2 - 2x + 1) = 0               factor out 12x 12x (x - 1) (x - 1) = 0               factor the x^2 factor 12x = 0 , x - 1 = 0 x = 0 , x = 1 restating the f' f'(x) = 12x^3 - 24 x^2 + 12x   find the second derivative f"(x) = 36x^2 - 48x + 12 f"(0) = 12 positive   that means there is a min at x = 0. f"(1) = 36(1) - 48(1) + 12 = 36 -48 + 12 = 0  at x = 1 there is a invertion point.

### 8x − 2 = y 5y − 12x = -3

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 − 12x = -3 5(8x − 2) − 12x = -3 40x - 10 - 12x = -3 28x - 10 = -3 28x - 10 + 10 = -3 + 10 28x = 7 28x/28 = 7/28 x = 7/28 x = 1/4 Now, solve for y. 8x − 2 = y 8(1/4) − 2 = y 2 - 2 = y 0 = y And check. 5y − 12x = -3 5(0) − 12(1/4) = -3 0 - 3 = -3 -3 = -3 Answer: x = 1/4, y = 0 Ordered pair: (1/4, 0)

### solve the linear system using any algebraic method 8x - 6y = 14 and 12x -9y = 18

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) 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)=3(14)_2*(12x-9y)=2(18) Multiply 3 by each term inside the parentheses. 3*(8x-6y)=42_2*(12x-9y)=2(18) Multiply 3 by each term inside the parentheses. (24x-18y)=42_2*(12x-9y)=2(18) Remove the parentheses around the expression 24x-18y. 24x-18y=42_2*(12x-9y)=2(18) Multiply 2 by each term inside the parentheses. 24x-18y=42_2*(12x-9y)=36 Multiply 2 by each term inside the parentheses. 24x-18y=42_(24x-18y)=36 Remove the parentheses around the expression 24x-18y. 24x-18y=42_24x-18y=36 Multiply the first equation by -1 to make the coefficients of y have opposite signs. -(24x-18y)=-(42)_24x-18y=36 Multiply -1 by the 42 inside the parentheses. -(24x-18y)=-42_24x-18y=36 Multiply -1 by each term inside the parentheses. (-24x+18y)=-42_24x-18y=36 Remove the parentheses around the expression -24x+18y. -24x+18y=-42_24x-18y=36 Add the two equations together to eliminate y from the system.  24x-18y=36_-24x+18y=-42_        =- 6 Since 0\$-6, there are no solutions. No Solution The system cannot be solved because it is inconsistent and has no intersection. The system cannot be solved because it is inconsistent.

### 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 easier (x-12)(x^5 - x^4 - 12x^3) = (x^6 - x^5 - 12x^4 - 12x^5 + 12x^4 + 144x^3) combin like terms x^6 - 13x^5 + 144x^3 (-12 + 12 cancels out the x^4) x^6 - 13x^5 + 144x^3 is your final answer! I really hope I did this right... I did this in my head... my method is correct, hoping my algebra is correct too was too lazy to get a paper + pencil so I worked it out in my head

### (12x + 2)(3x − 6)

(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  (12x + 2)(3x − 6)  = 0 then     12x  + 2 = 0       and also    3x  -  6  =  0 12x   =  -2   and           3x    = 6 x  =  -2/12      and    x   =  6/3 x   =  - 1/6    and  2

### 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"=4Ae^2x+4Be^2x=4p. Now for q. Since q"=0, q'=a, a constant, and q=ax+b. q"-4q must be equal to 12x so -4ax-4b=12x, making b=0 and a=-3. Therefore q=-3x and y=p+q=Ae^2x+Be^-2x-3x; y'=2Ae^2x-2Be^2x-3. y(0)=4=A+B; y'(0)=1=2A-2B-3, A-B=2. So 2A=6, A=3 and B=1. y=3e^2x+e^-2x-3x.

### write solution using set builder notation. 12x-25 is greater than or equal to 3x-5(x+6)

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}

### 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 36/3 x - 28/3 x = 1 - 88/3 8/3 x = -85/3 8x = -85 x = -85/8 x = -10 5/8 Plug that into equation 1 and solve for y. 7x - 6y = 22 7(-85/8) - 6y = 22 -595/8 - 6y = 22 -6y = 22 + 595/8 -6y = 22 + 74 3/8 -6y = 96 3/8 y = -16 1/16 Check by plugging both values into equation 1. 7x - 6y = 22 7(-85/8) - 6(-16 1/16) = 22 -74.375 + 96.375 = 22

### 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"=4Ae^2x+4Be^2x=4p. Now for q. Since q"=0, q'=a, a constant, and q=ax+b. q"-4q must be equal to 12x so -4ax-4b=12x, making b=0 and a=-3. Therefore q=-3x and y=p+q=Ae^2x+Be^-2x-3x; y'=2Ae^2x-2Be^2x-3. y(0)=4=A+B; y'(0)=1=2A-2B-3, A-B=2. So 2A=6, A=3 and B=1. y=3e^2x+e^-2x-3x.