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3 years ago

I'll give brainlist for this !!!!!!!!

Mathematics

1 answer:

castortr0y [4]3 years ago

4 0

Answer:

The first error in the bill is +8

because when two opposite signs of number are multiplied so the result is always negative.

so,

2 × -4 = -8

instead of 8 their should be -8.

hope this answer helps you dear...take care!

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A plane is an undefined term because it

igor_vitrenko [27]

No on can explain it but it has a definition.

8 0

3 years ago

Multiply: (x+4)(x^2+2x+3) <br> Simplify and write your answer in Standard Form.

Oxana [17]

Answer:

x^3+6x^2+11x+12

Step-by-step explanation:

(x+4)(x^2+2x+3)

=x(x^2+2x+3)+4(x^2+2x+3) [by multiplying with bith sides]

=x^3+2x^2+3x+4x^2+8x+12

=x^3+2x^2+4x^2+3x+8x+12

=x^3+6x^2+11x+12

(please mark me brainliest)

6 0

2 years ago

Find the volume of the solid.

dmitriy555 [2]

In Cartesian coordinates, the region (call it $R$ ) is the set

$R = \left\{(x,y,z) ~:~ x\ge0 \text{ and } y\ge0 \text{ and } 2 \le z \le 4-x^2-y^2\right\}$

In the plane $z=2$ , we have

$2 = 4 - x^2 - y^2 \implies x^2 + y^2 = 2 = \left(\sqrt2\right)^2$

which is a circle with radius $\sqrt2$ . Then we can better describe the solid by

$R = \left\{(x,y,z) ~:~ 0 \le x \le \sqrt2 \text{ and } 0 \le y \le \sqrt{2 - x^2} \text{ and } 2 \le z \le 4 - x^2 - y^2 \right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\sqrt2} \int_0^{\sqrt{2-x^2}} \int_2^{4-x^2-y^2} dz \, dy \, dx$

While doable, it's easier to compute the volume in cylindrical coordinates.

$\begin{cases} x = r \cos(\theta) \\ y = r\sin(\theta) \\ z = \zeta \end{cases} \implies \begin{cases}x^2 + y^2 = r^2 \\ dV = r\,dr\,d\theta\,d\zeta\end{cases}$

Then we can describe $R$ in cylindrical coordinates by

$R = \left\{(r,\theta,\zeta) ~:~ 0 \le r \le \sqrt2 \text{ and } 0 \le \theta \le\dfrac\pi2 \text{ and } 2 \le \zeta \le 4 - r^2\right\}$

so that the volume is

$\displaystyle \iiint_R dV = \int_0^{\pi/2} \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta \, dr \, d\theta \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} \int_2^{4-r^2} r \, d\zeta\,dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} r((4 - r^2) - 2) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \int_0^{\sqrt2} (2r-r^3) \, dr \\\\ ~~~~~~~~ = \frac\pi2 \left(\left(\sqrt2\right)^2 - \frac{\left(\sqrt2\right)^4}4\right) = \boxed{\frac\pi2}$