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3 years ago

What is the area of the figure below?

Mathematics

2 answers:

Alexus [3.1K]3 years ago

7 0

The answer is c 72m2

murzikaleks [220]3 years ago

6 0

96m^2 is your answer

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What is 1/10 the value of the digit 3 in 1,439.2

Artist 52 [7]

The value of the digit 3 is 30, so 3 is 1/10th.
Hope this helps

8 0

3 years ago

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2) Which expressions are equivalent to 1/3 - (-1/4). Select all that apply.

marshall27 [118]

What are all that apply?

8 0

3 years ago

Your dinner bill comes to $19.68. Estimate your total cost for dinner with a 5% tax and a 20% tip on the original bill. Justify

barxatty [35]

Estimate:
Bill = $20 approximately.
5% tax on $20 = 20/20=$1
20 % on ($20) = $4
Total cost = $20+1+4 = $25
Total bill = $25 approximately.

Exact calculations
bill = 19.68
tax 5% = 0.98
Tip 20% of original = 3.94
Total bill = 19.68+0.98+3.94 = 24.60

7 0

3 years ago

HELP PLEASE! 15 POINTS

Goshia [24]

Answer:

Its 58

Step-by-step explanation:

2+2+2+2+2+4+4+4+6+10+10=58 but the closest number to it is 56

8 0

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

3 years ago