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1 year ago

Add (x^3 -2x^2+4) + (2x^3+x^2-2) express in standard form

Mathematics

1 answer:

GalinKa [24]1 year ago

7 0

Answer:

Step-by-step explanation:

add both the equations then you will get

3x^3-x^2+2

but in the standard form there should be an x^1 term or x term since there is no term like that we write it as 0x

so the equation will be

3x^3-x^2+0x+2

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what is the slope of the line that passes through the points (-4,-4) and (-4,-9)? Write your answer in simplest form

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undefined

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3 years ago

22. Determine the value of c so that the line

tiny-mole [99]

Answer:

If the lines BC and DE are parallel, the value of c is c=-2

Step-by-step explanation:

We are given line segment BC with end points B(2, 2) and C(9,6) and line segment DE with endpoints D(c, -7) and E(5, -3).

Using slope formula: $Slope=\frac{y_2-y_1}{x_2-x_1}$ we can find point c

When 2 lines are parallel there slope is same.

So, Slope of line BC =Slope of Line DE

$\frac{y_2-y_1}{x_2-x_1}=\frac{y_2-y_1}{x_2-x_1}$

We have:

$x_1=2, y_1=2, x_2=9, y_2=6 \ for \ line \ BC \ and \\\x_1=c, y_1=-7, x_2=5, y_2=-3 \ for \ line \ DE \$

Putting values and finding c

$\frac{6-2}{9-2}=\frac{-3-(-7)}{5-c}\\ \frac{4}{7}=\frac{-3+7}{5-c} \\ \frac{4}{7}=\frac{4}{5-c} \\Cross \ multiply:\\4(5-c)=4*7\\20-4c=28\\-4c=28-20\\-4c=8\\c=\frac{8}{-4}\\c=-2$

So, If the lines BC and DE are parallel, the value of c is c=-2

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3 years ago

Add (x^3 -2x^2+4) + (2x^3+x^2-2) express in standard form​

Add (x^3 -2x^2+4) + (2x^3+x^2-2) express in standard form