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2 years ago

15 point reward + brainliest to correct answer!!

Mathematics

2 answers:

victus00 [196]2 years ago

7 0

Answer:

im here for the points

Step-by-step explanation:

lys-0071 [83]2 years ago

6 0

What’s the question that u need an answer to

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3 gallon of milk at $3.69 per gallon

Sonbull [250]

11.07 just do 4.69x 3

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3 years ago

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Expand and simplify 4(x-2)+7(x-1)

Zolol [24]

Answer:

11x-15

Step-by-step explanation:

4x-8+7x-7

11x-15

5 0

3 years ago

Someone plz help me with These questions :u have to solve for x and show work

irakobra [83]

Ok so the first one is
1.x=135
2.x=-23
3.x=3
4.x=70

5 0

3 years ago

What fraction is equivalent to 5/9

Mama L [17]

1. Any fraction that you multiply the numerator by the same number as the denominator.
Example:
1 x 2 = 2
-------------
2 x 2 = 4
2. The simplest form of 24/60 is 2/5 I believe.
3. 4/9
4. 23/8

Please vote me brainliest on this one too
thx

3 0

3 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

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

3 0

1 year ago