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3 years ago

My sister needs help with her homework. We tried 900 but it was not the right answer. Plz help!

Mathematics

2 answers:

katrin [286]3 years ago

8 0

Answer:

Ellie counted 888 plates

Step-by-step explanation:

Since Ellie counted 111 plates from 8 states, you only have to multiply 111 plates by the 8 states.( 111×8 = 888)

I hope this helps :)

damaskus [11]3 years ago

3 0

If your talking about estimating to the nearest hundred it would be 900 is your talking nearest ten it would be 890.

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Lelu [443]

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2 years ago

Read 2 more answers

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

Des makes 4 trays of granola bars.each tray holds 8 bars how many granola bars dose des make? Choose a strategy to solve.use num

iris [78.8K]

The answer is 32. You have four trays each have 8 bars so multiply 8 and 4.

8*4=32

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3 years ago

A cone has a volume of 226.08 cubic millimeters and a height of 6 millimeters. What is its

dezoksy [38]

Answer:

r≈5.36656314

Step-by-step explanation:

Since A=πr(r+ $\sqrt{h^2+r^2}$ ) so we get that 226.08/pi≈72 or 226.08/3.14=72 we then plug it in to get 72=r(r+ $\sqrt{36+r^2}$ ) so we get r≈5.36656314