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3 years ago

Can you guys help me find 1/3 + 5/6 please!

Mathematics

2 answers:

JulsSmile [24]3 years ago

8 0

Answer:

7/6

Step-by-step explanation:

1/3=2/6

2/6+5/6=7/6

Anettt [7]3 years ago

7 0

Answer:1.16666666667

Step-by-step explanation:

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Harrizon [31]

Answer:

62.8cubic units

Step-by-step explanation:

volume of a cylinder isπ×radius×radius×height.hope it helps.

4 0

2 years ago

I’ll mark brainliest !! Plz help

saveliy_v [14]

Answer:

3rd option

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Gail bought 1180 buttons to put on the shirts she she makes. She uses 6 buttons for each shirt. How shirts can Gail make with th

stepladder [879]

To answer this question, we need o divide.

Since she has 1180 buttons, we can divide that number by the amount of buttons it takes to make a shirt, 6.

1180/6=196

Hope this helps!

8 0

3 years ago

Read 2 more answers

Given the formula y = mx + b, what is the slope of a line, m, if x = 5, y = 7, and b =10?

anygoal [31]

The answer is:
C. The slope -3/5
Because -3/5 times 5 is -3 and when you combine -3 and 10 the answer is 7 and 7=7 (the y)

3 0

2 years 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:

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

7 0

2 years ago