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Serga [27]
3 years ago
9

Determine whether the sequence converges or diverges. If it converges, find the limit. [Hint:

Mathematics
1 answer:
Zepler [3.9K]3 years ago
6 0

(a) converges; consider the function <em>f(x)</em> = <em>a</em> ˣ, which converges to 0 as <em>x</em> gets large for |<em>a</em> | < 1. Then the limit is 2.

(b) converges; we have

4ⁿ / (1 + 9ⁿ) = (4ⁿ/9ⁿ) / (1/9ⁿ + 9ⁿ/9ⁿ) = (4/9)ⁿ × 1/(1/9ⁿ + 1)

As <em>n</em> gets large, the exponential terms vanish; both (4/9)ⁿ → 0 and 1/9ⁿ → 0, so the limit is 1.

(c) converges; we know ln(<em>n</em> ) → ∞ and arctan(<em>n</em> ) → <em>π</em>/2 as <em>n</em> → ∞. So the limit is <em>π</em>/2.

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True. The graph will shift down if we subtract a number from "b"

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If we don't change anything except for the y-intercept, then the graph should only move up or down.

Because we are decreasing the value of "b" the y intercept will decrease, dragging the graph of y down.

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A ticket for a school musical is $12. There is a 5% transaction fee if the ticket is purchased online. What is the total cost of
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Step-by-step explanation: First, we need to find the value of the transaction fee. Multiply 12 and 5%.

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3 years ago
Solve the polynomial inequality x^2-3x-10&gt;0
djverab [1.8K]

As x² - 3x - 10 factors, we can use the zero product property and pull it apart into factors. It factors because 2 and 5 have a difference of 3 and product of 10

x² - 3x - 10 > 0

(x - 5)(x + 2) > 0

To find where the inequality is true, we use a sign test. When x-5 > 0, it means x >5, and 5 is one of our sign test values. Similarly, -2 is the other. We can then make three intervals - A (less than -2), B (between -2 and 5), and C (more than 5).

<--------------- -2 ------------------ 5 -------------------------->

A B C

To do a sign test, we choose values to see what happens. First, test x = -5.

(-5 - 5)(-5 + 2) = (-10)(-3) = 30. Because 30 > 0, the inequality is true for x < -2. So interval A is true.

Next, test x = 0.

(0 - 5) (0 + 2) = (-5)(2) = -10. Because -10 < 0 is false, the inequality is false for x > -2 and x < 5. So interval B is false.

Last, test x = 10

(10 - 5) (10 + 2) = (5) (8) = 40. Because 40 > 0 is true, the inequality is true for x > 5. So interval C is true.

We put the true intervals together.

(-∞, -2) ∪ (5, ∞) is where this inequality is true,

{x : x < -2 or x > 5} is also how it's said in set notation.

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