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3 years ago

Indicate the equation of the given line in standard form, writing the answer in the equation box below.

Mathematics

1 answer:

yawa3891 [41]3 years ago

5 0

9514 1404 393

Answer:

x -3y = 4

Step-by-step explanation:

A plot of the points reveals the longer diagonal to be BD. (Also, point C is on that line.)

We can write the standard form equation of the line through (x1, y1) and (x2, y2) as ...

(y2 -y1)x -(x2 -x1)y = (y2 -y1)x1 -(x2 -x1)y1

For points (-2, -2) and (16, 4), this equation is ...

(4 -(-2))x -(16 -(-2))y = (4 -(-2))(-2) -(16 -(-2))(-2)

6x -18y = -12 +36

Dividing by the greatest common factor (6), we have the standard-form equation ...

x -3y = 4

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dybincka [34]

i) The given function is

$f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}$

The domain is

$(x-5)(x+4)^2\ne 0$

$(x-5)\ne0,(x+4)^2\ne 0$

$x\ne5,x\ne -4$

ii) For vertical asymptotes, we simplify the function to get;

$f(x)=\frac{(2x+1)}{(x+4)^2}$

The vertical asymptote occurs at

$(x+4)^2=0$

$x=-4$

iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

$2x+1=0$

$2x=-1$

$x=-\frac{1}{2}$

iv) To find the y-intercept, we substitute $x=0$ into the reduced fraction.

$f(0)=\frac{(2(0)+1)}{(0+4)^2}$

$f(0)=\frac{(1)}{(4)^2}$

$f(0)=\frac{1}{16}$

v) The horizontal asymptote is given by;

$lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0$

The horizontal asymptote is $y=0$ .

vi) The function has a hole at $x-5=0$ .

Thus at $x=5$ .

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

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butalik [34]

78
16 times 72 = 1152/(192/13) = 78

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RideAnS [48]

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