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how to multiply rational expression of (x^2 + xy / 15y^3) (35y^2 / 6x+6y) (12x / 10y^2

how to multiply rational expression

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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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multiply 2xy(x^2-xy+4) by applying the distributive property ...


multiply the following by applying the distributive property 2xy(x^2 ... to write an equivalent expression. A.) (3•y)•z B.) x + ... (x^2 + xy / 15y^3) (35y^2 / 6x ...
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y^2-16y+64=16 - page 2 - Jiskha Homework Help


What is the excluded value for the rational expression x^2 ... a= 3/2x-6=y 2. x+y=7 a= y=-x+7 3. -13+6y=x a= y=x/6+13/6 4. -2y ... the polynomial 6x^3 - 18x^2 + 12x ...
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Factor Polynomials - analyzemath.com


Factor Polynomials. ... Perfect cube x 3 + 3x 2 y + 3xy 2 + y 3 = (x + y) 3 11: ... As a practice, multiply (3x + 2)(3x - 1) ...
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5 Rational Expressions, Equations, and Functions - Ace ...


... b31x - y2 1x - y2x + yDIVIDING RATIONAL EXPRESSIONSTo divide by a rational expression, multiply ... 2 + 10y + 41125y 3 - 82 , 15y - 22. a3b3. x ... 12x 2x 2 - 2x ...
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(2x-3)(x^2+5x-3)=2x+10x^2-15x+9 *Is this correct.? - jiskha.com


Math(Please check my work.) (2x-3)(x^2+5x-3)=2x+10x^2-15x+9 *Is this correct.? this is a different problem... Here's how this one works out: .....x^2 + 5x - 3 ...
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Chapter 7 Rational Expressions and Functions - Ace ...


NThe LCD is fundamental propertyx (x - 2);546 Chapter 7 Rational Expressions ... 34x - 12x + 3=12x - 256x - 23x + 1x - 4=6x ... of a rational expression?Multiply or ...
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Systems of Equations - Problems & Answers - Math10.com


Systems of Equations - Problems & Answers. ... {array}{|l} 6x - y = 11 \\ 12x - 2y ... {x-1}{3} + \frac{5y+1}{2} = \frac{x+10y-8}{6} \\ \frac{(x+2)(5y-2)}{2} = 5 ...
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how to multiply rational expression of (x^2 + xy / 15y^3) (35y^2 / 6x+6y) (12x / 10y^2


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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.
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3x³-2x²-19x-6/3x+1

((3x^(3)-2x^(2)-19x-6)/(3x+1)) Factor the polynomial using the rational roots theorem. (((x+(1)/(3))(x+2)(x-3))/(3x+1)) To add fractions, the denominators must be equal.  The denominators can be made equal by finding the least common denominator (LCD).  In this case, the LCD is 3.  Next, multiply each fraction by a factor of 1 that will create the LCD in each of the fractions. (((x*(3)/(3)+(1)/(3))(x+2)(x-3))/(3x+1)) Complete the multiplication to produce a denominator of 3 in each expression. ((((3x)/(3)+(1)/(3))(x+2)(x-3))/(3x+1)) Combine the numerators of all expressions that have common denominators. ((((3x+1)/(3))(x+2)(x-3))/(3x+1)) Any number raised to the 1st power is the number. ((((1)/(3))(3x+1)(x+2)(x-3))/(3x+1)) Reduce the expression by canceling out the common factor of (3x+1) from the numerator and denominator. (((3x+1)(x+2)(x-3))/((3x+1))*(1)/(3)) Reduce the expression by canceling out the common factor of (3x+1) from the numerator and denominator. ((x+2)(x-3)*(1)/(3)) Multiply the rational expressions to get ((x+2)(x-3))/(3). (((x+2)(x-3))/(3)) Remove the parentheses around the expression ((x+2)(x-3))/(3). ((x+2)(x-3))/(3)
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how do you find the LCD of two rational expressions?

factoring
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How do you divide rational expressions?

Multiply the top and bottom by the common denominator of all the fractions. Doing this will get rid of the fractions within a fraction. I assume the % sign is supposed to mean ÷ x²  ---- -y ---------- 2y² ---- -x   x²  ---- (-xy) -y ---------- {multiplied top and bottom by common denominator} 2y² ---- (-xy) -x x³ ----- {cancelled and multiplied what was left} 2y³ www.algebrahouse.com  
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solve the rational expression 4/(z+3)=2/(z-2)

4 / (z + 3) = 2 / (z - 2) So multiply both sides by (z + 3) and (z - 2), then you get:   4(z - 2) = 2(z + 3) Now multiply out the brackets: 4(z - 2) = 2(z + 3) => 4z - 8 = 2z + 6 Now subtract '2z' from both sides of the equation to get all the 'z' terms on the same side: So, 4z - 8 = 2z + 6 becomes 4z - 2z - 8 = 2z  - 2z + 6 => 2z - 8 = 6 Now add 8 to both sides of the equation: So, 2z - 8 = 6 becomes 2z - 8 + 8 = 6 + 8 => 2z = 14 Now divide through by 2 to find the value of z: (2z)/2 = 14/2 => z = 7   So the solution is z = 7.             
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find a third degree polynomial equation with rational coefficients that has roots -4 and 6+i

A cubic equation always has three roots. These three roots are: 1) three real roots or, 2) 1 real root and two complex roots If one of the two complex roots is a + ib, then the other complex root is a - ib. We are given two of the roots as -4 and 6+i. Since one of the roots is complex and equals 6+i, the the other complex root is 6-i. Our three roots then are: -4, 6+i, 6-i. Our three solutions to the cubic equation are: x = -4, x = 6+i, x = 6-i. Which can be rewritten as: x + 4 = 0, x - (6+i) = 0, x - (6-i) = 0 Multiplying these together gives us the original cubic equation. (x + 4)(x - (6+i))(x - (6-i)) = 0 Multiplying this out, (x + 4)(x^2 - (6-i)x - (6+i)x + (6-i)(6+i)) = 0 (x + 4)(x^2 - 12x + 6^2 - i^2) = 0 (x + 4)(x^2 - 12x + 37) = 0 x^3 - 8x^2 - 11x +148 = 0  
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(2x^5y)^3 / 36x^8y^6

Simplify ——————— 36 Dividing exponents :  2.1    23   divided by   22   = 2(3 - 2) = 21 = 2 Equation at the end of step  2  : 2x 15y3 (—————— • x8) • y6 9 Step  3  : Multiplying exponential expressions :  3.1    x15 multiplied by x8 = x(15 + 8) = x23 Equation at the end of step  3  : 2x 23y3 —————— • y6 9 Step  4  : Multiplying exponential expressions :  4.1    y3 multiplied by y6 = y(3 + 6) = y9 Final result :   2x 23y9 —————— 9
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how to solve this word problem

-12x^2+52x-35 -12x^2+52x-35 = (-2x+7)(6x-5) (-2x+7)(6x-5) = -2x*6x -2x(-5) + 7*6x + 7(-5)= - 12x^2 + 10x + 42x - 35 = - 12x^2 + 52x - 35
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express in simplest form, using positive rational exponents: 3x^-1/3(x^1/3-4x^4/3)

3x^(-1/3) * (x^(1/3) - 4x^(4/3)) = 3(x^(1/3) - 4x^(4/3)) / x^(1/3) = 3(x^((1/3) - (1/3)) - 4x^((4/3) - (1/3))) = 3(x^0 - 4x^(3/3)) = 3(1 - 4x^1) = 3(1 - 4x) = 3 - 12x
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