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Serga [27]
2 years ago
9

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

Mathematics
1 answer:
Zepler [3.9K]2 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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Help me please I need it for math tomorrow
34kurt

Answer:

8?

Step-by-step explanation:

i can't see it

6 0
2 years ago
"the quotent of a number and 4" i dont wanna fail math
Zielflug [23.3K]

Answer:

N/4

Step-by-step explanation:

quotient is the same as division

since we don't know what the number is we use n as the variable

so n/4

Hope this helps

4 0
2 years ago
A sampling of hemlock trees revealed that 7 out of 190 trees are infected with insects. About how many of the 950 hemlock trees
nignag [31]

Answer:

About 35 trees in a state park are infected with insects

Step-by-step explanation:

we know that

7 out of 190 trees are infected with insects

so

using proportion

Find out about how many of the 950 hemlock trees in a state park are infected with insects

\frac{190}{7}=\frac{950}{x} \\\\x=7(950)/190\\\\x= 35

therefore

About 35 trees in a state park are infected with insects

3 0
3 years ago
three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal
Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

=     x + x + (x + \frac{6}{5}) = \frac{27}{4}

=     3x + \frac{6}{5} = \frac{27}{4}

=     3x = \frac{27}{4} - \frac{6}{5}

( let's take the LCM of 4 and 5 = 20

=     3x = \frac{135}{20} - \frac{24}{20}

=     3x = \frac{111}{20}

=       x = \frac{111}{20} ÷ \frac{3}{1} = \frac{111}{20} × \frac{1}{3} = \frac{37}{20}

So, the weight of the equal bags are \frac{37}{20} and the weight of the third bag ( heavy one ) is \frac{37}{20} + \frac{6}{5} = \frac{37}{20} + \frac{24}{20} = \frac{61}{20}

1st bag =     \frac{37}{20} kg

2nd bag =  \frac{37}{20} kg

3rd bag =   \frac{61}{20} kg

Hope it helps...

7 0
3 years ago
For breakfast, Sarah can choose eggs, granola or oatmeal as a main course, and orange juice or milk for a drink. Sarah says that
svlad2 [7]

Answer:

The explanation is given below

Step-by-step explanation:

Her error is that she multiplied the possibilities of her meal that involves eggs with milk, granola & oatmeal but the possibilities involves orange juice with egg and granola & oatmeal

Now

If she choose milk so she can select eggs, granola, and oatmeal so there are 3 possibilities also with the orange juice there is 3 possibilities

So, the sample space is

= 3 + 3

= 6 outcomes  

4 0
2 years ago
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