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4 years ago

Divide x^3 -3x^2 by x-2 . Step 1 - Fill in the missing number:

Mathematics

2 answers:

vagabundo [1.1K]4 years ago

8 0

Answer:

Step-by-step explanation:

First let us write the given polynomial as in descending powers of x with 0 coefficients for missing items

F(x) = x^3-3x^2+0x+0

We have to divide this by x-2

Leading terms in the dividend and divisor are

x^3 and x

Hence quotient I term would be x^3/x=x^2

x-2) x^3-3x^2+0x+0(x^2

x^3-2x^2

Multiply x-2 by x square and write below the term and subtract

We get

x-2) x^3-3x^2+0x+0(x^2

x^3-2x^2

---------------

-x^2+0x

Again take the leading terms and find quotient is –x

x-2) x^3-3x^2+0x+0(x^2-x

x^3-2x^2

---------------

-x^2+0x

-x^2-2x

Subtract to get 2x +0 as remainder.

x-2) x^3-3x^2+0x+0(x^2-x-2

x^3-2x^2

---------------

-x^2+0x

-x^2+2x

-------------

-2x-0

-2x+4

------------------

-4

Thus remainder is -4 and quotient is x^2-x-2

pantera1 [17]4 years ago

4 0

Answer:

its 2

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

Graph and label the image of the figure below after a dilation by a factor of 1/2.

neonofarm [45]

Answer:

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

see graph below

Explanation:

Given:

The image of a quadrilateral on a coordinate plane

To find:

The coordinates of the new image after dilation of 1/2 have been applied to the original image.

Then graph the coordinates

First, we need to state the coordinates of the original image:

M = (3, -2)

F = (4, -2)

L = (1, -5)

W = (5, -5)

Next, we will apply a scale factor of 1/2:

$\begin{gathered} Dilation\text{ rule:} \\ (x,\text{ y\rparen}\rightarrow(kx,\text{ ky\rparen} \\ where\text{ k = scale factor} \\ \\ scale\text{ factor = 1/2} \\ M^{\prime}\text{ = \lparen}\frac{1}{2}(3),\text{ }\frac{1}{2}(-2)) \\ M^{\prime}\text{ = \lparen}\frac{3}{2},\text{ -1\rparen} \\ \\ F\text{ = \lparen}\frac{1}{2}(4),\text{ }\frac{1}{2}(-2)) \\ F^{\prime}\text{ = \lparen2, -1\rparen} \end{gathered}$ $\begin{gathered} L\text{ = \lparen}\frac{1}{2}(1),\text{ }\frac{1}{2}(-5)) \\ L^{\prime}\text{ = \lparen}\frac{1}{2},\text{ }\frac{-5}{2}) \\ \\ W\text{ = \lparen}\frac{1}{2}(5),\text{ }\frac{1}{2}(-5)) \\ W^{\prime}\text{ = \lparen}\frac{5}{2},\text{ }\frac{-5}{2}) \end{gathered}$

The new coordinates:

M' (3/2, -1), F' (2, -1), L' (1/2, -5/2), W' (5/2, -5/2)

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

Plotting the coordinates:

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