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OlgaM077 [116]
4 years ago
5

Find the horizontal asymptote

Mathematics
1 answer:
eimsori [14]4 years ago
5 0

Answer:

Correct answer:  h = 3/5  or y = 3/5

Step-by-step explanation:

Given:

h(x) = (3x² - 9x - 4) / (5x² - 4x + 8)

h = lim x -> ∞ (3x² - 9x - 4) / (5x² - 4x + 8) the numerator and denominator will be divided by x² and get

h = lim x -> ∞ (3 - 9(1/x) - 4 (1/x²)) / (5 - 4(1/x) + 8 (1/x²)

The terms 1/x and 1/x²  when x strives ∞ they strives 0

x -> ∞ ⇒ 1/x -> 0  and x -> ∞ ⇒ 1/x² -> 0

h = (3 - 0 - 0) / (5 - 0 + 0)

Horizontal asymptote is h = 3/5 or y = 3/5

God is with you!!!

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If area of a circle measures 16/25 pi cm square, what is the circumference of the circle in terms of pi
pishuonlain [190]
A = pi*r^2

(16/25)*pi = pi*r^2

16/25 = r^2

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r = 4/5


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7 0
3 years ago
Which members are in the sample
Anna [14]

Answer:

20, 26, 35, 18

Step-by-step explanation:

So starting at row 129, we look at the sequence two-digits at a time without overlapping.  If that number is between 01 and 43, then they get selected.

The first two digits are 20.  That fits between 01 and 43, so that member gets selected.

Next, we have 26.  That also fits.

After that we have 64.  Nope, too high.

98 and 44 are also too high.

35 fits though.  So does 18.

So the members that get selected are 20, 26, 35, 18.

5 0
4 years ago
Why do the factor not have to be prime?
dangina [55]
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4 years ago
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50 POINTS!!! Word problem.
oksano4ka [1.4K]
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Then plug in the value for d into the second equation.
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3 years ago
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