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olga55 [171]
3 years ago
9

Encuentren dos números cuyo cociente sea -1 y que la resta del primero menos el segundo sea -3/2

Mathematics
1 answer:
miskamm [114]3 years ago
4 0

Answer:

\dfrac{3}{4}\ \text{and}\ \dfrac{-3}{4}

Step-by-step explanation:

Find two numbers whose quotient is -1 and the subtraction of the first minus the second is -3/2

Let two numbers are x and y. When they are divided the quotient is -1. So,

\dfrac{x}{y}=-1

It can also be written as :

x = -y ....(1)

The subtraction of first minus the second is -3/2 .

x-y=\dfrac{-3}{2}\ .....(2)

Use equation (1) in (2),

-y-y=\dfrac{-3}{2}\\\\-2y=\dfrac{-3}{2}\\\\y=\dfrac{3}{4}

From equation (1) : x=\dfrac{-3}{4}

Hence, two numbers are \dfrac{3}{4}\ \text{and}\ \dfrac{-3}{4}.

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Carter divided the polynomial 6x2 + 4x + 3 by the monomial 2x using long division, and got an answer of 3x + 2 + . His work is s
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Answer:

C. Both Carter and Demi are correct and would get the same answer.

Step-by-step explanation:

Here, the given polynomial is : P(x) =6x^2 + 4x + 3

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Now, according to the given question:

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Now, Carter divide P(x) by Q (x) with LONG DIVISION METHOD.

If we divide P(x) by Q(x) by Long division, we get:

\frac{P(x)}{Q(x)}  = \frac{(6x^2 + 4x + 3)}{(2x)}  = 3x +  2 + \frac{3}{2} x = R(x)

Also, if Demi divides each term of P(x) by The given monomial (2x) , we get:

\frac{P(x)}{Q(x)}  = \frac{(6x^2 + 4x + 3)}{(2x)}  = \frac{6x^2}{2x}  + \frac{4x}{2x} +\frac{3}{2x} \\= 3x +  2 + \frac{3}{2}  = R(x)

⇒In both the case, R(X)Quotient is SAME.

Hence, Both Carter and Demi are correct and would get the same answer.

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