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2 years ago

An airplane flies 315 km in 1 hour 45 minutes. Calculate the average speed.

Mathematics

1 answer:

valentina_108 [34]2 years ago

4 0

Answer:

3 km per minute

Step-by-step explanation:

speed = distance / time

315km/105 minutes

= 3 km per minute

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Regis has a bag with 8 tiles numbered 1 through 8. He randomly draws one bag without looking. Which of the following describes a

Daniel [21]

He has a likely outcome of 1/8

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2 years ago

What number is 100,000,000 more than 5,438,724,022

Slav-nsk [51]

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3 years ago

What are all the ratios that are equivalent to 8:6

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Answer:

4:3, 16:12, 24:18, and many more

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3 years ago

Which expression will simplify to 1?

JulsSmile [24]

Let us check each Option :

$\mathsf{First\;Option : \left(\dfrac{m + 9}{m - 9}\right)\left(\dfrac{m + 9}{m - 9}\right)}$

$\mathsf{\implies \left(\dfrac{m + 9}{m - 9}\right)^2\;\neq\;1}$

$\mathsf{Second\;Option : \left(\dfrac{m + 9}{m - 9}\right)\left(\dfrac{m - 9}{m + 9}\right)}$

$\mathsf{\implies 1}$

$\mathsf{Third\;Option : \left(\dfrac{m + 9}{m - 9}\right)\left(\dfrac{9 + m}{9 - m}\right)}$

$\mathsf{\implies \left(\dfrac{m + 9}{m - 9}\right)\left(\dfrac{m + 9}{-(m - 9)}\right)}$

$\mathsf{\implies-\left(\dfrac{m + 9}{m - 9}\right)\left(\dfrac{m + 9}{m - 9}\right)}$

$\mathsf{\implies-\left(\dfrac{m + 9}{m - 9}\right)^2\;\neq\;1}$

$\mathsf{Fourth\;Option : \left(\dfrac{m + 9}{m - 9}\right)\left(\dfrac{9 - m}{9 + m}\right)}$

$\mathsf{\implies \left(\dfrac{m + 9}{m - 9}\right)\left(\dfrac{-(m - 9)}{9 + m}\right)}$

$\mathsf{\implies -\left(\dfrac{m + 9}{m - 9}\right)\left(\dfrac{m - 9}{9 + m}\right)}$

$\mathsf{\implies -1\;\neq\;1}$

Answer : Option (2)

7 0

3 years ago

Can someone help me with this? **SHOW WORK PLEASE**

Valentin [98]

FYI, that L in the denominator is factorial.

This is some pretty serious stuff for high school.

My inclination is that the answer is A, which is the definition of e. Let's see if we can show this.

Let's write

$\displaystyle f(n) = \left(1 + \dfrac 1 n \right)^n$

Let's expand this with the binomial expansion

$\displaystyle f(n) =\sum_{k=0}^n {n \choose k} \dfrac{1}{n^k}$

$\displaystyle f(n) =\sum_{k=0}^n \dfrac{n!}{k!(n-k)!} \cdot \dfrac{1}{n^k}$

$\displaystyle f(n) =\sum_{k=0}^n \dfrac{n(n-1)\cdots(n-k+1)}{k! \, n^k}$

Let's focus on when n is really big and on the ks that are relatively small, which make up the bulk of the sum as the terms get small rapidly.

Then that numerator n(n-1)···(n-k+1) ≈ n^k as all the factors are about n.

$\displaystyle f(n)\approx \sum_{k=0}^n \dfrac{n^k}{k!\, n^k}$

$\displaystyle f(n)\approx \sum_{k=0}^n \dfrac{1}{k!}$

OK, we showed for large n this is approximately true, and it will be exactly true in the limit, so we choose

Answer: A

3 0

3 years ago