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2 years ago

Help pls I don’t understand this my teacher didn’t teach this

Mathematics

2 answers:

wlad13 [49]2 years ago

5 0

Answer:

Step-by-step explanation:

X=60

aev [14]2 years ago

4 0

Answer:

x = 60

Step-by-step explanation:

The tip of the triangle's angle is 100 (b/c 180-80)

x + 100 + 20 = 180

x + 120 = 180

x = 60

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Anuta_ua [19.1K]

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1 year ago

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Simplify 3(tan^2 theta - sec^2 theta)

Nesterboy [21]

Answer:

sin ^2(x)

Step-by-step explanation:

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2 years ago

How do you solve this using substitution method 17x+7y=20 Y=20x-42

Nutka1998 [239]

Answer:

17 x + 7 y = 20

y =20x - 42

17 x + 7(20x - 42) = 20

17x + 140x - 294 = 20

157 x = 20 + 294

157x ÷ 157 = 314 ÷ 157

x = 2

y = 20x - 42 

y = 20 (2) - 42

y = 40 - 42

y = -2

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1 year ago

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:

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

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2 years ago

What is 12% of 3500?

kakasveta [241]

Answer:

29,166.6667

Step-by-step explanation:Divide 12% which will turn into 0.12 then divide that with 3500.

6 0

2 years ago

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