Select the exclusion that fits best with this problem. 3x^2y^2-6xz/6x^2yz

Mathematics

2 answers:

katrin2010 [14]3 years ago

4 0

Answer:

Option B is correct .i.e., x , y , z = 0 must be excluded.

Step-by-step explanation:

Given Expression:

$\frac{3x^2y^2-6xz}{6x^2yz}$

We have to find value of x , y and z which are to be excluded.

Value of variable of a expression are excluded for which expression does not exist.

Given expression is of fraction,

so, where denominator is going to be zero those value going to be excluded.

Here Denominator = 6x²yz

Denominator can not be equal to 0

So to find point where it become zero we put 6x²yz equal to 0

6x²yz = 0

⇒ x² = 0 or y = 0 or z = 0

Therefore, Option B is correct .i.e., x , y , z = 0 must be excluded.

andreyandreev [35.5K]3 years ago

3 0

For this case we have the following expression:
3x ^ 2y ^ 2-6xz / 6x ^ 2yz
From here, we must exclude all values that make the denominator equal to zero.
We have then:
6x ^ 2yz = 0
Therefore, the results are:
x = 0
y = 0
z = 0
Answer:
The exclusion that fits best with this problem is:
x, y, z = 0

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Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antid

romanna [79]

Answer:

1) F(x)= $\frac{2x^{5} }{5}+x^{4}-\frac{x^{2} }{2}+C$

2) $f(x)=\frac{8x^5}{5}-\frac{3x^4}{2}+\frac{4x^3}{3}+D$

Step-by-step explanation:

Applying the antiderivation rules:

F(x)= $\frac{2x^{4+1} }{4+1}+4\frac{x^{3+1} }{3+1}-\frac{x^{1+1} }{1+1}+C$

F(x)= $\frac{2x^{5} }{5}+x^{4}-\frac{x^{2} }{2}+C$

Checking by differentitation we have:

$F'(x)=2*\frac{5x^{4} }{5}+4*\frac{4x^3}{4}+2*\frac{x}{2}\\F'(x)=2x^4+4x^3-x=f(x)$

Which is demonstrated.

2) To find f we must antiderivate twice:

$f'(x)=\int\limits {(32x^3-18x^2+8x)} \, dx\\f'(x)=8x^4-6x^3+4x^2+C\\f(x)=\int\limits( {8x^4-6x^3+4x^2)} \, dx\\ f(x)=8\frac{x^5}{5}-6\frac{x^5}{5}+4\frac{x^3}{3}+D\\ f(x)=\frac{8x^5}{5}-\frac{3x^4}{2}+\frac{4x^3}{3}+D$