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3 years ago

Fernando built a robot to filter air and water efficiently. He expects the robot to filter more than 512 liters

Mathematics

1 answer:

Mumz [18]3 years ago

4 0

Answer:

Khan Academy

The robot uses the expected amount of energy, but it doesn't filter the expected amount of air and water.

Step-by-step explanation:

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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

3 years ago

Pleassssssseeeeeee heeelllllppppppppp

olga2289 [7]

Answer: 1. point A

2. 90 degrees

3. 90 degrees clockwise

Step-by-step explanation:

BRAINLIEST??!?!

6 0

4 years ago

Find the midpoint of A and B where A has coordinates (-5, 3) and B has coordinates (3,-1)

yaroslaw [1]

Answer:

(-1, 1)

Step-by-step explanation:

Midpoint formula: (x₁ + x₂/2 , y₁ + y₂/2)

Substitute the values into the formula.

(-5 + 3/2 , 3 - 1/2)

(-2/2 , 2/2)

(-1,1)

Therefore, the midpoint is (-1, 1).

5 0

3 years ago

X^(2)-4x-17=0 Please show work step by step

Dimas [21]

Step-by-step explanation:

It seems here that they are asking us to solve for x

to do this we first need to factor

Since we can't factor this using the normal method we can instead do this

x^2 -4x-17=0

add 4 to both sides as it is a perfect square

x^2 - 4x + 4 = 17 +4

(x-2)^2 = sqrt 21

x-2 = ± 4.58

x -2 = 4.58 x-2 = -4.58

x= 6.58 x=-2.58

Or just say x=2+√21 or x=2−√21

7 0

3 years ago

Helppp plsssss someone D:

zimovet [89]

〽Hola User_______________

⭐Here is Your Answer..!!!

______________________

↪Cartesian Coordinates of the system....!!

↪X = rcos⊙ and Y = rsin⊙

↪since here Radius (r) = 10 units and Theta (⊙) = 225°

↪there fore substituting we get as ..

↪X = 10 ( cos 225° ) = 10 * -1/ root 2 = -7.071

↪Y = 10 ( sin 225° ) = 10 * -1/root2 = = -7.071

↪the cordinates are as follows

↪(x,y) = (-7.071, -7.071 )

4 0

3 years ago