What is the following sum? Assume x greater-than-or-equal-to 0 and y greater-than-or-equal-to 0StartRoot x^2y^3 EndRoot +2 StartRoot x^3y^4 EndRoot +xy StartRoot y EndRoot

Assume x greater-than-or- equal-to 0 and y ... 0 StartRoot x squared y cubed EndRoot + 2 StartRoot x cubed y Superscript 4 Baseline EndRoot + x y StartRoot y EndRoot.

Step-by-step explanation:

Option (b) is correct.

The sum of given expression  \sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}x2y3+2x3y4+xyy is  2xy\sqrt{y}+2xy^2\sqrt{x}2xyy+2xy2x

Step-by-step explanation:

Given:Expression \sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}x2y3+2x3y4+xyy

We have to find the sum of the given expression and choose the correct from the given options.

Consider the given expression  \sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}x2y3+2x3y4+xyy .

\sqrt{x^2y^3}x2y3 can be written as \sqrt{x^2y^2y}=xy\sqrt{y}x2y2y=xyy

Also, 2\sqrt{x^3y^4}2x3y4 can be written as 2\sqrt{x^3y^4}=2\sqrt{x^2x(y^2)^2}=2xy^2\sqrt{x}2x3y4=2x2x(y2)2=2xy2x

Now, the given expression becomes,

x2y3+2x3y4+xyy

=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}=xyy+2xy2x+xyy .

Now, adding like term, terms having same variable with same degree.

=2xy\sqrt{y}+2xy^2\sqrt{x}=2xyy+2xy2x .

Thus, The sum of given expression  \sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}x2y3+2x3y4+xyy is  2xy\sqrt{y}+2xy^2\sqrt{x}2xyy+2xy2x

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation: