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

step-by-step explanation:

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)