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nataly862011 [7]
3 years ago
11

Identify the horizontal asymptote of f(x) =x2+5x-3/4x-1

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
5 0

since the numerator is x² + 5x - 3, and therefore has a degree of 2, whilst the denominator, 4x¹ - 1, has a degree of 1, therefore, there's no horizontal asymptote.

recall, we only get a horizontal asymptote if the denominator's expression degree is equals or greater than that of the numerator's.

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I really need help with this is due on Friday pls help with steps!
olchik [2.2K]

Answer:

1. 11

2. 7

3. 4

4. 10

.

5. 17.2

6. 2.2

7. 3.3

8. 7.9

Step-by-step explanation:

When calculating square root, you always need to find a number multiplied by itself twice that gives the number

Example: \sqrt{64}

What number multiplied by itself twice gives 64? Number 8! because 8 × 8 = 64

Now cube root is a little bit different. This time, you will need to find a number multiplied by itself three times that gives that number.

Example: \sqrt{125}

What number multiplied by itself three times gives 125? Number 5! because 5 × 5 × 5 = 125

Hope this helps!

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A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has four identi
Tcecarenko [31]

Answer:

a. 0.1536

b. 0.9728

Step-by-step explanation:

The probability that a component fails, P(Y) = 0.2

The number of components in the system = 4

The number of components required for the subsystem to operate = 2

a. By binomial theorem, we have;

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) is given as follows;

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

The probability that exactly 2 last longer than 1,000 hours, P(Y = 2) = 0.1536

b. The probability that the system last longer than 1,000 hours, P(O) = The probability that no component fails + The probability that only one component fails + The probability that two component fails leaving two working

Therefore, we have;

P(O) = P(Y = 0) + P(Y = 1) + P(Y = 2)

P(Y = 0) = \dbinom{4}{0} × 0.2⁰ × 0.8⁴ = 0.4096

P(Y = 1) = \dbinom{4}{1} × 0.2¹ × 0.8³ = 0.4096

P(Y = 2) = \dbinom{4}{2} × 0.2² × 0.8² = 0.1536

∴ P(O) = 0.4096 + 0.4096 + 0.1536 = 0.9728

The probability that the subsystem operates longer than 1,000 hours = 0.9728

4 0
2 years ago
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