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Serga [27]
3 years ago
12

-2x + 3 < 5 Show your work/steps

Mathematics
1 answer:
s2008m [1.1K]3 years ago
5 0
<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ -2x + 3 < 5

Subtract 3 from both sides:

-2x < 2

Divide both sides by -2 to isolate x:

x = -1

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ <u>ʜᴀɴɴᴀʜ</u> ♡

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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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3 years ago
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postnew [5]

Answer:

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Step-by-step explanation:

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