All categories

3 years ago

Does anyone know how to do this?

Mathematics

1 answer:

GalinKa [24]3 years ago

3 0

Diagonals of parallelogram ABCD are AC and Bd. P is point where diagonals crossing

You might be interested in

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

2 years ago

I struggled with these as u can see, but pls answer all 3 questions and show how you set all the problems up (ill be giving brai

olga55 [171]

Answer:

1+1-2 its 2

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Emily thinks the perfect tomato sauce has 8 cloves of garlic in every 500{ mL} of sauce. Raphael's tomato sauce has 12 cloves of

igomit [66]

For perfect solution we need to have :- 8 colves in 500ml solution

500 ml has 8 cloves

1 ml has 8 / 500 cloves

100ml has 8 / 500 * 100 = 8/5 = 1.6 cloves

Raphael's mixture

900 ml has 12 cloves

1 ml has 12 / 900 cloves

100 ml has 12 / 900 * 100 = 12 / 9 = 1.33 cloves

so concentration of garlic in 100 ml solution of Raphael's solution is less than Emily's solution so it is not garlicky enough. ( option B)

5 0

2 years ago

Read 2 more answers

If y varies inversely as x and y=16 when x=4, find y when x=3

iren [92.7K]

Answer:

21.33 to nearest hundredth

Step-by-step explanation:

Inverse Variation is y = k / x where k is a constant.

Plug in the given values to find k:-

16 = k/4

k = 4*16 = 64 so the equation of variation is y = 64/x.

So when x = 3, y = 64/3 = = 21.33 to nearest hundredth.

6 0

3 years ago

HI HELP ITS A QUIZZZ!

Virty [35]

B. Its suggesting something bigger than -5

4 0

3 years ago

Does anyone know how to do this?​

Does anyone know how to do this?