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Inessa [10]
2 years ago
5

What is the horizontal asymptote of f(x) = (2x+1)/ (x^2-4x+3) +2

Mathematics
1 answer:
evablogger [386]2 years ago
6 0

Answer:

x=3 and x=1

Step-by-step explanation:Horizontal asymptote will occur at x=3 and x=1.

You have to factorize the denominator and find the zeroes of the denominator.  That is how you find out the horizontal asymoptote.

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Please answer this ??
Vadim26 [7]
The answer is 24.5. Just plug the numbers in
7 0
2 years ago
A storekeeper has two kinds of flour, one selling for 65 cents per pound, and the other selling for 95 cents per pound. How many
Murrr4er [49]

Answer:

The amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 pounds

Step-by-step explanation:

Let

x ----> amount of 65 cents per pound flour to use

y----> amount of 95 cents per pound flour to use

we know that

x+y=100 ----> y=100-x ----> equation A

0.65x+0.95y=0.83(100) ----> equation B

substitute equation A in equation B

0.65x+0.95(100-x)=0.83(100)

Solve for x

0.65x+95-0.95x=83

0.95-0.65x=95-83

0.30x=12

x=40\ lb

Find the value of y

y=100-40=60\ lb

therefore

The amount of 65 cents per pound flour to use is 40 pounds and the amount of 95 cents per pound flour to use is 60 pounds

4 0
3 years ago
Read 2 more answers
How do you write 0 .25 in a fraction?
tatuchka [14]
0.25 in a fraction would be 1/4
4 0
3 years ago
Read 2 more answers
Solve:
riadik2000 [5.3K]

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Let's solve ~

\qquad \sf  \dashrightarrow \:  \dfrac{2x - 1}{5}  =  \dfrac{x -  2}{2}

\qquad \sf  \dashrightarrow \: 2(2x - 1) = 5(x - 2)

\qquad \sf  \dashrightarrow \: 4x - 2 = 5x - 10

\qquad \sf  \dashrightarrow \: 5x - 4x = -2 + 10

\qquad \sf  \dashrightarrow \: x = 8

Value of x is 8

8 0
2 years ago
Read 2 more answers
Is X-1 a factor of x^5-1?
Musya8 [376]

Answer: No

Explanation:

According to factor theorem, if f(x)=0 then x is a factor of the given function or equation.

As x-1 is a factor

We equate x-1=0

x=1

Substituting in x^5-1, we have 1^5-1 =1-1=0.

Hence, it's a factor.

When coming to x^5+1, it would become 1^5+1=1+1=2

So x-1 isn't a factor of x^5+1.

4 0
2 years ago
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