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zalisa [80]
3 years ago
13

Hi there can anyone help me with this question please

Mathematics
1 answer:
Umnica [9.8K]3 years ago
8 0
Range is 1 because they all go up by 1
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Answer:

Too much

Step-by-step explanation:

I did the math myself

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Ten white socks and five red socks are in a drawer. If two socks are selected at random, what is the probability that both socks
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Evaluate the following expression when a = 5.<br> 4a2 + a-3
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4(5)2+5-3
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<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.

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Step-by-step explanation:

\sqrt{2} is an irracional number it has infinetely many digits

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2(\sqrt{2}+1 )

a decimal aproach is

4.828...

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