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3 years ago

Perform the indicated operation to write given polynomial in standard form: (x^3+2x^2 − 3x − 1) +(4 − x − x^3)

Mathematics

1 answer:

Rzqust [24]3 years ago

5 0

Answer:

$\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-4x+3$

Step-by-step explanation:

To write any polynomial in standard form, you look at the degree of each term. You then write each term in order of degree, from highest to lowest, left to write.

To add the polynomials you need to:

Remove parentheses

$\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=x^3+2x^2-3x-1+4-x-x^3$

Group like terms

$\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=x^3-x^3+2x^2-3x-x-1+4$

Add similar elements

$\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-3x-x-1+4\\\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-4x-1+4\\\left(x^3+2x^2-3x-1\right)+\left(4-x-x^3\right)=2x^2-4x+3$

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Find the distance between the point (-2, -3) and the line with the equation y=4-2/3x

otez555 [7]

Answer:

$D = \frac{25\sqrt{13}}{13}$

Step-by-step explanation:

Given

$y = 4 - \frac{2}{3}x$

$(x_1,y_1) = (-2,-3)$

Required

Determine the distance

$y = 4 - \frac{2}{3}x$

Write the above equation in standard form:

$Ax + By + C = 0$

So, we have:

$\frac{2}{3}x+y - 4 = 0$

By comparison:

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

The distance is calculated using:

$D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Where:

$(x_1,y_1) = (-2,-3)$

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

This gives:

$D = \frac{|\frac{2}{3} * (-2) + 1 * (-3) - 4|}{\sqrt{(\frac{2}{3})^2 + 1^2}}$

$D = \frac{|-\frac{4}{3} -3 - 4|}{\sqrt{\frac{4}{9} + 1}}$

Take LCM

$D = \frac{|\frac{-4-9-12}{3}|}{\sqrt{\frac{4+9}{9}}}$

$D = \frac{|\frac{-25}{3}|}{\sqrt{\frac{13}{9}}}$

$D = |\frac{-25}{3}|/\sqrt{\frac{13}{9}}$

$D = \frac{25}{3}/\sqrt{\frac{13}{9}}$

Split the square root

$D = \frac{25}{3}/\frac{\sqrt{13}}{\sqrt{9}}$

Change / to *

$D = \frac{25}{3}*\frac{\sqrt{9}}{\sqrt{13}}$

$D = \frac{25}{3}*\frac{3}{\sqrt{13}}$

$D = \frac{25}{\sqrt{13}}$

Rationalize

$D = \frac{25}{\sqrt{13}} * \frac{\sqrt{13}}{\sqrt{13}}$

$D = \frac{25\sqrt{13}}{13}$

Hence, the distance is:

$D = \frac{25\sqrt{13}}{13}$

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3 years ago

1. Which of the following in the process of finding the factors of the

Sidana [21]

Answer:

We cannot say it's different Difference ot of Two Cubes because 2d2 is not not cube it's square. and 8d is not a cube.

We cannot say Difference of Two Squares because only first term 2d2 has a square.

It is not a Perfect Square Trinomials because Perfect Square Trinomials appears as ax2 + bx + c, but the given ter doesn't follow this.

A. Common Monomial Factor can be regarded as a variable, or more than one variable that that is present in the polynomial terms.

Example is 4x2 + 16x,

If factorize we have 4x(x + 4) with monomer of 4x and polynomial of x + 4). that cannot be factorized into lower polynomial.

Hence 2d2-8d can be factor as 2d(d-4) where 2d is the monomer and (d-4) cannot be factorize into lower degree polynomial.

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3 years ago

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Answer:

A. 3/8 of an hour

Step-by-step explanation:

if she is not a slacker

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3 years ago

What is 11/15= 8/15-w

victus00 [196]

Answer: w= 1/5

Step-by-step explanation:

11/15=8/15 -w

-W= 8/15 -11/15

-w= 8-11/15 =-3/15

-w=-3/15

w=3/15= 1/5

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forsale [732]

Answer:

y = 1/2x + 1

Step-by-step explanation:

Step 1: Find slope

(1-0)/(0+2) = 1/2

Step 2: Write equation

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