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3 years ago

Multiply, (2x^4-3y^2)(x^2+5y^4)

Mathematics

2 answers:

Ad libitum [116K]3 years ago

7 0

10x^4y^4+2x^6-15y^6-3x^2y^2

aleksandr82 [10.1K]3 years ago

4 0

To answer this question, I use FOIL (First, Outside, Inside, Last)

First: 2x^4 x x^2 = 2x^6
Outside: 2x^4 x 5y^4 = (10xy)^8
Inside: -3y^2 x x^2 = (-3yx)^4
Last: -3y^2 x 5y^4 = -15y^6

Now you just add all the answers up :)

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Integrate G(x,y,z)equalsz over the parabolic cylinder yequalszsquared, 0less than or equalsxless than or equals2, 0less than

yKpoI14uk [10]

I gather you're supposed to compute the integral of $G(x,y,z)=z$ over a surface $S$ that is the part of the parabolic cylinder $y=z^2$ with $0\le x\le2$ and $0\le z\le\frac{\sqrt{15}}2$ .

We can parameterize $S$ by

$\vec s(x,z)=x\,\vec\imath+z^2\,\vec\jmath+z\,\vec k$

with the given constraints on $x$ and $z$ . Take the normal vector to $S$ to be

$\vec s_x\times\vec s_z=-\vec\jmath+2z\,\vec k$

so that the surface element is

$\mathrm dS=\|\vec s_x\times\vec s_z\|\,\mathrm dx\,\mathrm dz=\sqrt{1+4z^2}\,\mathrm dx\,\mathrm dz$

Then in the integral, we have

$\displaystyle\iint_Sz\,\mathrm dS=\int_0^2\int_0^{\sqrt{15}/2}z\sqrt{1+4z^2}\,\mathrm dz\,\mathrm dx=\boxed{\frac{21}2}$

5 0

3 years ago

Factor the expression completely over the complex numbers.<br><br> y^4+12y^2+36

Brums [2.3K]

Y⁴ + 12y² + 36
Now factorize the expression
y⁴ + 6y² + 6y² + 36
= y²(y² + 6) + 6(y² + 6)
= (y² + 6) (y² + 6)
<span>Now 6 is not the perfect square and according to rule, binomial can not be factored as the difference of two perfect squares.
</span>so multiply both.
(y² + 6)² is the answer.

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3 years ago

Read 2 more answers

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babunello [35]

Answer: 38 is your answer hope this helped

plz make brainly

Step-by-step explanation:

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kupik [55]

Answer:

1 5/7 hours.

Step-by-step explanation:

1/3+1/4 = 7/12

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3 years ago

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aleksandrvk [35]

The answer is b=2 because yea

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