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Mathematics
x
2 – 3xy – 5xy – 7y
2 + 4x
2 + 8y
2
1 answer
I don't mind helping to find answers, but I resent having to provide the questions as well.
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Is this correct now after i revised it
Multiply. 3xy^2(5xy-4x+6y) 3xy^2(5xy)=15x^2y^3 3xy^2(-4x)=-12x^2y^2 3xy^2(6y)=18xy^3 so i
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Multiply: (3xy + 2y) by 5xy
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What is the simplest form of the expression (5x + 3xy + 4y) + (4x – 2xy – 2y)?
A. x + xy + 2y B. x^2 + xy + 2y C. 9x + 5xy +
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no thoughts on any of these?
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Which expression is equivalent to
(3xy+2xz+yz)+(2xy-4xz+5yz)? 5xy-6xz-4yz 5xy-2xz+6yz 5xy-2xz-4yz 5xy-6xz-6yz
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The correct expression is 5xy-2xz+6yz.
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Which expression is equivalent to `\left(3xy\ +\ 2xz\ +\ yz\right)\ +\ \left(2xy\ -\ 4xz\ +\ 5yz\right)?` 5xy\ -\ 6xz\ -4yz 5xy\
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The expression can be simplified by combining like terms: \begin{align*} \left(3xy + 2xz + yz\right)
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Which expression is equivalent to `\left(3xy\ +\ 2xz\ +\ yz\right)\ +\ \left(2xy\ -\ 4xz\ +\ 5yz\right)?`
5xy-6xz-4yz 5xy-2xz-6yz
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To simplify the expression, combine like terms by adding or subtracting their coefficients: (3xy +
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Which expression is equivalent to `\left(3xy\ +\ 2xz\ +\ yz\right)\ +\ \left(2xy\ -\ 4xz\ +\ 5yz\right)?`
5xy-6xz-4yz 5xy-2xz+6yz
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To simplify the expression, combine like terms. The like terms in the expression are: (3xy + 2xy) =
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whats the answer to 3xy(2x^2+5xy-7y^2
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To find the answer to the expression 3xy(2x^2+5xy-7y^2), we need to apply the distributive property
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Which is equivalent to the following expression? (3x^2+5xy−2y^2)+(x^2+3xy−y^2)
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The expression (3x^2+5xy-2y^2)+(x^2+3xy-y^2) is equivalent to (3x^2 + x^2) + (5xy + 3xy) + (-2y^2 -
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the following steps are used to rewrite the polynomial expression
step 1. x(x-5y)+2y(x-5y) step 2.x^(2)-5xy+2yx-10y^(2) step 3.
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Step 1: Distributive Property Step 2: Combining like terms Step 3: Combining like terms Step 4:
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the following steps are used to rewrite the polynomial expression
step 1. x(x-5y)+2y(x-5y) step 2.x^(2)-5xy+2yx-10y^(2) step 3.
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The given polynomial expression can be rewritten through the following steps: Step 1: Distribute x
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