Answer:
are the roots.
Step-by-step explanation:

Considering the expression

Solving





Solving

= 
![\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D)
= 
![\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^m}=a^{\frac{m}{n}}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D)



So,

Similarly,

Therefore,
are the roots.
Keywords: roots, expression
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<h3>☂︎ Answer :- </h3>
<h3>☂︎ Solution :- </h3>
- LCM of 5 , 18 , 25 and 27 = 2 × 3³ × 5²
- 2 and 3 have odd powers . To get a perfect square, we need to make the powers of 2 and 3 even . The powers of 5 is already even .
In other words , the LCM of 5 , 18 , 25 and 27 can be made a perfect square if it is multiplied by 2 × 3 .
The least perfect square greater that the LCM ,
☞︎︎︎ 2 × 3³ × 5² × 2 × 3
☞︎︎︎ 2² × 3⁴ × 5²
☞︎︎︎ 4 × 81 × 85
☞︎︎︎ 100 × 81
☞︎︎︎ 8100
8100 is the least perfect square which is exactly divisible by each of the numbers 5 , 18 , 25 , 27 .
Plug in the 1/2 where the a’s are answer is -3
Answer:
The perimeter of a quarter of a circle with radius 2 inches is 7.1 inches.
Step-by-step explanation:
Given : A quarter of a circle with radius labeled 2 in.
Quarter of circle means fourth part of a circle (as shown in image by shaded part).
We have to find the perimeter of this shaded quarter of circle.
Perimeter is the sum of the dimension of the given figure.
Perimeter of circle is 
Since, we are given Quarter circle so its perimeter 
Perimeter of given figure = 
Perimeter of given figure = 
Thus, the perimeter of a quarter of a circle with radius 2 inches is 7.1 inches.