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2 years ago

I need help it's so confusing!

Mathematics

1 answer:

jasenka [17]2 years ago

5 0

Try looking up the answer

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The sum of two polynomials is -yz to the second power-3z to the second power-4y+4. If one of the polynomials is y-4yz to the sec

balu736 [363]

Answer:

- 5yz^2 - 3z^2 - 5y + 7

Step-by-step explanation:

The polynomial is:

$-yz^2-3z^2-4y+4$ ( 1 )

this polynomial is the result of the sum of two polynomials, one of them given by

$y-4yz^2-3$ ( 2 )

if you find to know what is the other polynomial, you can take the difference between the polynomial (1) and (2). It is necessary to take the difference between simmilar terms, as follow:

- yz^2 - 3z^2 - 4y + 4

- (4yz^2 + y - 3 )

- yz^2 - 3z^2 - 4y + 4

- 4yz^2 - y + 3

- 5yz^2 - 3z^2 - 5y + 7

hence the other polynomial is - 5yz^2 - 3z^2 - 5y + 7

5 0

3 years ago

What is the product? Enter your answer as a fraction, in simplified form, in the box. 38⋅(−36)

valina [46]

If you multiply 38*(-36) you would get -1368. And to convert that into a fraction it could be -1368/1, -13680/10, etc.

8 0

3 years ago

Read 2 more answers

Let V be the volume of the solid obtained by rotating about the y-axis the region bounded by y=√x and y=x^2. Find V either by sl

Ede4ka [16]

Answer:

The volume of the solid is $3\pi/10$

Step-by-step explanation:

In this case, the washer method seems to be easier and thus, it is the one I will use.

Since the rotation is around the y-axis we need to change de dependency of our variables to have $f(x)\rightarrow f(y)$ . Thus, our functions with $y$ as independent variable are:

$x=\sqrt{y}\\ x=y^2$

For the washer method, we need to find the area function, which is given by:

$A=\pi\cdot [(\rm{outer\ radius)^2 -(\rm{inner\ radius)^2 ]$

By taking a look at the plot I attached, one can easily see that for a rotation around the y-axis the outer radius is given by the function $x=\sqrt{y}$ and the inner one by $x=y^2$ . Thus, the area function is:

$A(y)=\pi\cdot [(\sqrt{y} )^2-(y^2)^2]\\A(y)=\pi\cdot (y-y^4)$

Now we just need to integrate. The integration limits are easy to find by just solving the equation $\sqrt(y)=y^2$ , which has two solutions $y=0$ and $y=1$ . These are then, our integration limits.