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3 years ago

Identify the GCF of 10x4y3 − 5x3y2 + 20x2y. 10x2y 10x3y 5x2 5x2y

Mathematics

2 answers:

Allisa [31]3 years ago

7 0

Answer: $5x^2y$

Step-by-step explanation:

Given polynomial : $10x^4y^3-5x^3y^2 + 20x^2y$

We can rewrite the terms given in the above polynomial as :-

$10x^4y^3=5\times2\timesx^4y^3\\\\-5x^3y^2=5\times-1\timesx^3y^2\\\\20x^2y=5\times4\timesx^2y$

The highest common factor of numerical coefficients = 5

The highest common power of x =2

The highest common power of y =1

Therefore, the greatest common factor (GCF) of $10x^4y^3-5x^3y^2 + 20x^2y= 5x^2y$

Nata [24]3 years ago

6 0

10x^4y^3 - 5x^3y^2 + 20x^2y....first, find the GCF of ur coefficients (numbers)....it is 5...now find the lowest exponents of x, which is x^2...and now find ur lowest exponent of y, which is y

so ur GCF is : 5x^2y

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Read 2 more answers

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Answer:

a) $\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b) $\large \mathbb{R}^2-\{(0,0)\}$

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Step-by-step explanation:

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

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On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$

$\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

The domain of definition of F is

$\large \mathbb{R}^2-\{(0,0)\}$

i.e., all the plane X-Y except the (0,0)

Here we want to find all the points such that

$\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)$

where k is a real number other than 0.

But this means

$\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x$

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

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