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3 years ago

What are two steps are necessary to put this equation into standard form x^2 - 3x + 27 =8x- 3

Mathematics

1 answer:

Wittaler [7]3 years ago

7 0

Answer:

you need to bring the 8x over then the -3

Step-by-step explanation:

X^2-3x+27=8x-3

X^2-11x+27=-3

x^2-11x+24

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3 years ago

Is the sqrt of 1 - sin^2 theta = cos theta true? If so, in which quadrants does angle theta terminate?

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Step-by-step explanation:

Here,

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3 years ago

A line includes the points (-39,49) and (0,0). What is its equation in slope-intercept form?

malfutka [58]

Answer:

The equation in the slope-intercept form will be:

$y=-\frac{49}{39}x+0$

Step-by-step explanation:

Given the points

(-39,49)

(0,0)

Finding the slope between the points using the formula

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

$\left(x_1,\:y_1\right)=\left(-39,\:49\right),\:\left(x_2,\:y_2\right)=\left(0,\:0\right)$

$m=\frac{0-49}{0-\left(-39\right)}$

$m=-\frac{49}{39}$

We know that the point-slope of the line equation is

$y-y_1=m\left(x-x_1\right)$

substituting $m=-\frac{49}{39}$ and (-39,49) in the equation

$y-y_1=m\left(x-x_1\right)$

$y-49=\frac{-49}{39}\left(x-\left(-39\right)\right)$

Now writing the equation in slope-intercept form

$y=mx+b$

where m is the slope and b is the y-intercept

$y-49=\frac{-49}{39}\left(x-\left(-39\right)\right)$

$y-49=\frac{-49}{39}\left(x+39\right)$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}$

$y-49=-\frac{49}{39}\left(x+39\right)$

$\mathrm{Add\:}49\mathrm{\:to\:both\:sides}$

$y-49+49=-\frac{49}{39}\left(x+39\right)+49$

$y=-\frac{49}{39}x+0$ ∵ $y=mx+b$

Where

$m=-\frac{49}{39}$ and the y-intercept i.e. $b=0$

Therefore, the equation in the slope-intercept form will be:

$y=-\frac{49}{39}x+0$

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3 years ago