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3 years ago

I'll give you all stars if you answer my questions.

Mathematics

1 answer:

k0ka [10]3 years ago

4 0

I would say E, systematic. It isn't random, nor convenience, or clustered. It also isn't stratified.

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I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

8 0

2 years ago

Which of the equations are true identities? A. N ( n − 2 ) ( n + 2 ) = n 3 − 4 n B. ( x + 1 ) 2 − 2 x + y 2 = x 2 + y 2 + 1

ICE Princess25 [194]

Answer:

Both A and B are true identities

Step-by-step explanation:

A. N ( n − 2 ) ( n + 2 ) = n 3 − 4 n

We need to show that (left-hand-side)L.H.S = R.H.S (right-hand-side)

So,

n ( n − 2 ) ( n + 2 ) = n(n² - 2²) (difference of two squares)

= n³ - 2²n (expanding the brackets)

= n³ - 4n (simplifying)

So, L.H.S = R.H.S

B. ( x + 1 )² − 2x + y² = x² + y² + 1

We need to show that (left-hand-side)L.H.S = R.H.S (right-hand-side)

So,

( x + 1 )² − 2x + y² = x² + 2x + 1 - 2x + y² (expanding the brackets)

= x² + 2x - 2x + 1 + y² (collecting like terms)

= x² + 1 + y²

= x² + y² + 1 (re-arranging)

So, L.H.S = R.H.S

So, both A and B are true identities since we have been able to show that L.H.S = R.H.S in both situations.

7 0

3 years ago

The difference of the square of a number and three times the sum of the number and 7 is more than the product of that number and

Pachacha [2.7K]

Answer:

x2 - 3(x + 7) > 15x

x2 - 3x - 21 > 15x

6 0

4 years ago

Please help me with #2!!!!

Natali [406]

D. 5

It is what it is

8 0

3 years ago

What is the equation of this line?

dezoksy [38]

Answer:

A. y = 2x - 3

Explaination:

It can't be negative because then the slope would be going downward from left to right, this line is going upwards. Also, the steepness of the slope proves it is a slope of 2 because a slope of 1/2 isn't very steep. You can also count 1 unit over and look how much the line changes to find the slope.

7 0

3 years ago

Read 2 more answers