Answer:
<span>The irrational number between 5.5 and 5.2 is square root of 30. </span>
Answer:
250
Step-by-step explanation:
you have to use the fact that the big triangle and the small triangle are similar and write the proportions
we took half of the side because of ΔABD and ΔBD..
AB= (125*10)/10=125
is AB =125 then AC is duble AC= 2*125=250
Answer:
A
Step-by-step explanation:
Find AB using the Law of Sine:

Thus:

Multiply both sides by sin(40)


AB = 31.0056916 ≈ 31.0 cm (nearest tenth)
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,

This is a rhombus and in any rhombus, the diagonals intersects in the middle and they are perpendicular:
So all 4 triangles are right triangles and the sides are the hypotenuses.
1st) Calculate the sides: hypotenuse² = 3² + 4² = 25, and hypotenuse = 5
The area of each right triangle is (4 x 3)/2 = 6 units²
And the area of the 4 right triangles = 4 x 6 = 24 init²