Answer:
The area of the paralellogram is 24 square units.
Step-by-step explanation:
Geometrically speaking, the area of the parallelogram has an equation equivalent to the area formula for a rectangle, that is:
(1)
Where:
- Area.
- Base.
- Height.
The base and height of the parallelogram are, respectively:
Base


Height


Then, the area of the parallelogram is:


The area of the paralellogram is 24 square units.
Answer:
350 girls
Step-by-step explanation:
So,
The secret to solving problems with ratios is to find the value of one unit.
5:7 = 12 units total
To find one unit, divide the total number of students by the total number of units.
600/12 = a
Simplify
50/1 = a
50 = a
The value of each unit is 50.
Now, multiply the units by the numbers in the ratio.
50(5) = b
250 = boys
50(7) = x
350 = x
There are 350 girls.
Answer:
A
Step-by-step explanation:
given the sequence above using using two as a first term the only formula fits the sequence above is A
Step-by-step explanation:

Given expression is

To, evaluate this limit, let we simplify numerator and denominator individually.
So, Consider Numerator

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.
So, using Sum of n terms of GP, we get


Now, Consider Denominator, we have

can be rewritten as

![\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%20%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B%7B%5Cdfrac%7Bn%20-%201%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7Bn%20-%202%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D)
![\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%20%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D)
Now, Consider

So, on substituting the values evaluated above, we get
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%5Cdfrac%7B%20%7Bn%7D%5E%7Bn%7D%20%20-%201%7D%7B1%20-%20%20%5Cdfrac%7B1%7D%7Bn%7D%20%7D%7D%7B%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%20%7Bn%7D%5E%7Bn%7D%20%20-%201%7D%7B%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%20%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20-%20%5Cdfrac%7B1%7D%7B%20%7Bn%7D%5E%7Bn%7D%20%7D%20%5Cbigg%5D%7D%7B%7Bn%7D%5E%7Bn%7D%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B%5Cbigg%5B1%20-%20%5Cdfrac%7B1%7D%7B%20%7Bn%7D%5E%7Bn%7D%20%7D%20%5Cbigg%5D%7D%7B%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
![\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}](https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20%20%3D%20%20%5C%3A%20%5Cdisplaystyle%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Cfrac%7B1%7D%7B%5Cbigg%5B1%20%2B%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%5Cbigg%5B1%20-%20%7B%5Cdfrac%7B2%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%2B%20%20-%20%20-%20%20-%20%20%2B%20%5Cbigg%5B%7B%5Cdfrac%7B1%7D%7Bn%7D%5Cbigg%5D%7D%5E%7Bn%7D%20%5Cbigg%5D%7D%20)
Now, we know that,
![\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}](https://tex.z-dn.net/?f=%5Cred%7B%5Crm%20%3A%5Clongmapsto%5C%3A%5Cboxed%7B%5Ctt%7B%20%5Cdisplaystyle%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Cbigg%5B1%20%2B%20%5Cdfrac%7Bk%7D%7Bx%7D%20%5Cbigg%5D%5E%7Bx%7D%20%20%3D%20%20%7Be%7D%5E%7Bk%7D%7D%7D%7D%20)
So, using this, we get

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have





Hence,

Answer:
A
Step-by-step explanation: