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3 years ago

Find if divergent/convergent: a(sub n)=(n^2/sqr(n^3+4n)), if convergent, find the limit.

1 answer:

4 0

We are given the equation an = (n^2/ sqrt(n^3+4n)) and asked to determine if the function is divergent or convergent. In this case, we find the limit of the function as n approaches infinity.

an = (n^2/ sqrt(n^3+4n))
lim (n to infinity ) = infinity / infinty: ;indeterminate

Using L'hopitals rule, we derive
lim (n to infinity ) = 2 n / 0.5* ( n^3+4n) ^-0.5 * (3 n2 +4) : infinity / infinity

again, we derive

lim (n to infinity ) = 2 (0.25) (( n^3+4n) ^-0.5))*(3 n2 +4) / 0.5* ( 6n + 4) :infinity / infinity

again,
lim (n to infinity ) = 2 (0.25) (6n + 4) / 0.5* ( 6)* 0.5 (( n^3+4n) ^-0.5))

this goes on and the function is divergent

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Yusuf walked 4 miles per hour for 0.7 hours. How far did Yusuf walk?

DIA [1.3K]

Answer:

Yusuf walked 2.8 miles.

Step-by-step explanation:

r = 4 miles / hour

t = 0.7 hours.

d = r*t

d = 4 * 0.7

d = 2.8 miles

4 0

3 years ago

Verify that:

Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

$\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)$

Note that cot(x) = cos(x) / sin(x). Hence:

$\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)$

Multiply:

$\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

$\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Split:

$\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Simplify:

$\csc(x)-\sin(x)=\csc(x)-\sin(x)$

Problem 2)

We want to verify that:

$\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}$

Square:

$\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}$

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

$\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out the sin²(x) from the denominator:

$\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (perfect square trinomial):

$\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

$\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (difference of two squares):

$\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out a negative from the first factor in the denominator:

$\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Cancel:

$\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Distribute the negative into the numerator. Therefore:

$\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}$

3 0

3 years ago

2 1 MLD

mash [69]

Profits = 3.009(number of pizzas) - 77.131
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4 0

3 years ago

What is the mad of 7, 9, 6, and 6

Alborosie

The Mean Absolute Deviation of 7,9,6,6 is 1 hope it helps <3

3 0

3 years ago

Please answer as soon as you can.

madam [21]

Ratio is 4 please like!

6 0

3 years ago