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3 years ago

What is the relationship between two coplanar lines that are perpendicular to the same line?

Mathematics

2 answers:

Charra [1.4K]3 years ago

6 0

Answer:

I'm sorry for this but I'll help you if you help me?

grigory [225]3 years ago

3 0

Answer:

If two lines are perpendicular to the same line, then the two lines are parallel, and if two lines are parallel, then their slopes are equal

Step-by-step explanation:

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Square root of 244 by division method

navik [9.2K]

Answer:

The square root of 244 is 15.62 ( up-to 2 decimal places)

Step-by-step explanation:

Following is the division method for square root of 244 with rules as follows:

Make pairs of the given digit starting from right (unit place). They will be calculated one by one as formed in pairs/singles.
Digit added for divisor and quotient will be same at each step.
Each next dividend will be formed by subtracting product of divisor and quotient from first digit/pair.
Each next divisor will be formed by taking two times of previous quotient and annexing it with suitable digit that gives equal to or less than the dividend.

Following solution gives square root of 244:

I hope it will help you!

7 0

3 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

3 years ago

Brody worked 8.5 hours at Food Lion this week and made $78.20. Based on the number of hours Brody worked and how much he was pai

Lynna [10]

Answer:

1 hour = $9.2

Step-by-step explanation:

The question says that:

Brody makes $78.20 in 8.5 hours.

So,

8.5 hours = $78.20

We need to find 1 hour pay.

So, divide both sides of the above equation by 8.5

8.5 / 8.5 hours = $78.20 / 8.5

1 hour = $9.2

$\rule[225]{225}{2}$

Hope this helped!

6 0

3 years ago

The Dot plot shows the salaries for the employees that two small companies before a new company head is hired on each company se

Dmitry_Shevchenko [17]

Answer:

Step-by-step explanation:

8 0

3 years ago

Find X (Show Your Work) WILL MARK BRAINLIEST!

Rudiy27

Answer:

its 8 cause 2 times x is 40 40+-40= 0 so 2 times 4 is 8

Step-by-step explanation:

6 0

3 years ago