Answer:
Option D. ![\sqrt[4]{\frac{3x^{2}}{2y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B3x%5E%7B2%7D%7D%7B2y%7D%7D)
Step-by-step explanation:
![\sqrt[4]{\frac{24x^{6}y}{128x^{4}y^{5}}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B24x%5E%7B6%7Dy%7D%7B128x%5E%7B4%7Dy%5E%7B5%7D%7D%7D)
![\sqrt[4]{(\frac{24}{128})\times (\frac{x^{6}}{x^{4}})\times (\frac{y}{y^{5}})}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%28%5Cfrac%7B24%7D%7B128%7D%29%5Ctimes%20%28%5Cfrac%7Bx%5E%7B6%7D%7D%7Bx%5E%7B4%7D%7D%29%5Ctimes%20%28%5Cfrac%7By%7D%7By%5E%7B5%7D%7D%29%7D)
= ![\sqrt[4]{(\frac{3}{16})\times {(x)^{6-4}}\times{(y)^{1-5}}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%28%5Cfrac%7B3%7D%7B16%7D%29%5Ctimes%20%7B%28x%29%5E%7B6-4%7D%7D%5Ctimes%7B%28y%29%5E%7B1-5%7D%7D%7D)
= ![\sqrt[4]{(\frac{3}{16})\times x^{2}y^{-4}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%28%5Cfrac%7B3%7D%7B16%7D%29%5Ctimes%20x%5E%7B2%7Dy%5E%7B-4%7D%7D)
= ![\sqrt[4]{\frac{3}{(2)^{4}}\times x\times y^{-4}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B3%7D%7B%282%29%5E%7B4%7D%7D%5Ctimes%20x%5Ctimes%20y%5E%7B-4%7D%7D)
= ![\sqrt[4]{(3\times x^{2)\times (\frac{y^{-1}}{2})^{4}}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%283%5Ctimes%20x%5E%7B2%29%5Ctimes%20%28%5Cfrac%7By%5E%7B-1%7D%7D%7B2%7D%29%5E%7B4%7D%7D%7D)
= ![\frac{y^{-1}}{2}\sqrt[4]{3x^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7By%5E%7B-1%7D%7D%7B2%7D%5Csqrt%5B4%5D%7B3x%5E%7B2%7D%7D)
= ![\sqrt[4]{\frac{3x^{2}}{2y}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B3x%5E%7B2%7D%7D%7B2y%7D%7D)
Option D.
is the correct answer.
well It never could be less than both numbers.
the LCM of 2 numbers shd either be greater than the 2 numbers .... or the answer should be equal to either one of the numbers ...
lets see some examples to prove this conclusion:
now, lets find the LCM of 2 and 6 ... so if we do it ... we get the answer 6 .. which is ..as we said equal to one of the numbers
another example : 5 and 7 ... in this ... if we do it .. .we get the LCM as 37 ... which is greater than both the numbers
another case where both the numbers are same ....
where we need to find the LCM of 7 and 7 ... the answer will be 7 itself .....
Answer:
x = 14
Step-by-step explanation:
Extend line AB so that it intersects ray CE at point G. Then angles BGC and BAD are "alternate interior angles", hence congruent.
The angle at B is exterior to triangle BCG, and is equal to the sum of the interior angles at C and G:
138 = (376 -23x) +(x^2 -8x)
Subtracting 138 and collecting terms we have ...
x^2 -31x +238 = 0
For your calculator, a=1, b=-31, c=238.
__
<em>Additional comment</em>
You will find that the solutions to this are x = {14, 17}. You will also find that angle BCE will have corresponding values of 54° and -15°. That is, the solution x=17 is "extraneous." It is a solution to the equation, but not to the problem.
For x=14, the marked angles are A = 84°, C = 54°.
<h3>
Answer:</h3>
A=160 m
<h3>
Solution:</h3>
- The formula for the Area of a rectangle is:
- A=LW
- In the above formula, L stands for the length and W stands for the width.
- We are given:
- L=16 m
- W=10 m
- We are asked to find the area of the rectangular pool.
- The operation that we're going to use here is:
- Multiplication.
- Multiply:
- A=16*10
- A=160 m
Hope it helps.
Do comment if you have any query.
Answer:

Step-by-step explanation:
