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Tpy6a [65]
3 years ago
14

Helpppppp pleaseeeeeeeeeeee

Mathematics
2 answers:
Nesterboy [21]3 years ago
4 0
Ya is shebought 6 errs seeds sk
Lyrx [107]3 years ago
3 0

Answer:

She bought 6 erasers.

Step-by-step explanation:

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3 years ago
Find Horizontal and Vertical asymptotes for y=(x^2 -9)/(4x^2 +1)
Gnom [1K]

Answer: Horizontal asymptote is \dfrac{1}{4} and vertical asymptotes are \pm \dfrac{1}{2i}

Step-by-step explanation:

Since we have given that

y=\dfrac{x^2-9}{4x^2+1}

We need to find the horizontal and vertical asymptotes.

Since vertical asymptotes will occur where the denominator becomes zero.

So, here denominator is 4x^2+1

Now,

4x^2+1=0\\\\4x^2=-1\\\\x^2=\dfrac{-1}{4}\\\\x=\pm \dfrac{1}{2i}

And the horizontal asympototes will occur when the coefficient of higher degree of numerator is divided by coefficient of higher degree of denominator.

y=\dfrac{1}{4}

Hence, horizontal asymptote is \dfrac{1}{4} and vertical asymptotes are \pm \dfrac{1}{2i}

8 0
3 years ago
Please help me with this question
Lorico [155]
This is what I got I hope it helps

8 0
3 years ago
Read 2 more answers
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