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3 years ago

10

PLS HELP! i will mark you the brainliest!!!

1 answer:

forsale [732]3 years ago

8 0

1. v = 4.5

2. Whatever variable = 6.3

3. g = 0.4

4. a = 17.4

Hope it helps! :)

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ROUND 90998.0014483 TO THE NEAREST TEN. HELP!

djverab [1.8K]

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4 years ago

Does the following represent direct variation? If yes, find the constant variation

svlad2 [7]

y/x =k

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3 years ago

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3 years ago

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Studentka2010 [4]

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b

Step-by-step explanation:

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3 years ago

<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7Bx-5%7D%7B2x%5E%7B2%7D-5x-3%20%7D" id="TexFormula1" title="f(x) = \f

Vedmedyk [2.9K]

i) The given function is

$f(x)=\frac{x-5}{2x^2-5x-3}$

The factored form is

$f(x)=\frac{x-5}{(x-3)(2x+1)}$

The domain are the values of x for which the function is defined.

$(x-3)(2x+1)\ne 0$

$(x-3)\ne0,(2x+1)\ne 0$

$x\ne3,x\ne-\frac{1}{2}$

ii) To find the vertical asymptotes, equate the denominator to zero.

$(x-3)(2x+1)=0$

$(x-3)=\ne0,(2x+1)=0$

$x=3,x=-\frac{1}{2}$

iii) To find the roots, equate the numerator to zero.

$x-5=0$

The root is $x=5$

iv) To find the y-intercept, put $x=0$ into the function.

$f(0)=\frac{0-5}{(0-3)(2(0)+1)}$

$f(0)=\frac{-5}{(-3)(1)}$

$f(0)=\frac{5}{3}$

The y-intercept is $\frac{5}{3}$

v) The horizontal asymptote is given by;

$lim_{x\to \infty}\frac{x-5}{2x^2-5x-3}=0$

The horizontal asymptote is $y=0$

vi) The function is not reducible. There are no holes.

vii) The given function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

3 0

3 years ago

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