3 years ago

How do i find the slant asymptope of x^3/(2(x+2)(x-4) aka x^3/(2x^2-8)

1 answer:

6 0

I’m really sorry I just don’t know I guess I’m. It smart enough try to look it up and wait for someone else to answer sorry again :(

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Solve x2 + 10x + 12 = 36 for x.

tankabanditka [31]

X=2 my friend
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{c(1) = 3/16<br> {c(n)=c(n−1)⋅4<br> <br><br> what is the 3rd term in this sequence?

mote1985 [20]

Answer:

$\large\boxed{c(3)=3}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}c(1)=\dfrac{3}{16}\\\\c(n)=c(n-1)\cdot 4\end{array}\right$

Put n = 2 and next n = 3 to the recursive formula:

$c(2)=c(2-1)\cdot4=c(1)\cdot4\to c(2)=\dfrac{3}{16}\cdot4=\dfrac{3}{4}\\\\c(3)=c(3-1)\cdot4=c(2)\cdot4\to c(3)=\dfrac{3}{4}\cdot4=3$

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The explicit formula for the nth term of an arithmetic sequence is an = a1 + (n - 1) • d. What is the simple formula correspondi

BlackZzzverrR [31]

= -10 + (n - 1)*4
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= 4n - 14

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3 years ago

If $10,500 is deposited in a compound interest account paying 2.99% interest annually, how much will be in the account after 5 y

algol [13]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$10500\\ r=rate\to 2.99\%\to \frac{2.99}{100}\dotfill &0.0299\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases} \\\\\\ A=10500\left(1+\frac{0.0299}{1}\right)^{1\cdot 5}\implies A=10500(1.0299)^5\implies A\approx 12166.47$

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3 years ago

Can someone please answer this with step by step i really need help thank you

Gre4nikov [31]

Answer:

40°

I hope it will be useful.

Step-by-step explanation:

Angles on the same side of the transversal are equal.

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