Which equation would help you find the missing side length in the right triangle shown?

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

Genber wiil evaluate an 8th degree polynomial in x at x=10 using the remainder theorem and synthetic division.how many coefficie

lisabon 2012 [21]

An 8th-degree polynomial needs 9 terms that involve
x⁸, x⁷, ..., x¹, and x⁰.

x=10 implies that (x-10) is a factor of the polynomial according to the Remainder theorem.

Let the polynomial be of the form
f(x) = a₁x⁸ + a₂x⁷ + a₃x⁶ +a₄x⁵ + a₅x⁴ + a₆x³ + a₇x² + a₈x + a₉

The first few lines of the synthetic division are

10 | a₁ a₂ a₃ a₄ a₅ a₆ a₇ a₈ a₉ ( the first row has 9 coefficients)

-----------------------------------------
a₁

Answer:
The first row has 9 coefficients.

8 0

2 years ago

Circle the following 2 terms that correspond with the domain of a function:

Nitella [24]

Answer:

x-values

independent variable

Step-by-step explanation:

x (the independent variable) is the domain

y (the dependent variable) is the range

8 0

3 years ago

Read 2 more answers

Write which numbers,if any,are solutions of the inequality. N+9<15 N=? 3,5,6,7,9

mash [69]

The answer is
n=6
n<15-9
n<6

8 0

2 years ago

What’s the proof for this

KIM [24]

Well... Basically, you should prove this by SSS property(side-side-side). It's fair to say that the length of a side is equal to itself so The line that cuts through the rectangle is a side for both rectangles. Thus because of the given, all the sides of the triangles are equal to one another. This is a very important trick for geometry(I remember using it a lot).
Hope this helps!

7 0

2 years ago