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3 years ago

.can someonee help me as soon as possible

Mathematics

1 answer:

Nezavi [6.7K]3 years ago

7 0

Answer:

y=8

x=20

Step-by-step explanation:

8y+6=70

5x-10=90

because(language persian or farsi)

چون که شکل متوازی الاضلاع است پس دو زاویه مجاور مکمل هستند

همچنین چون feوgh دو خط موازی هستند xhg میشود 70 درجه

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Four and seventy-two thousandths

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Answer:

4.072

Step-by-step explanation:

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What times 5 equals 75

saveliy_v [14]

15 the answer is 15 because 15 x 5 = 75

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Solve for x.

Ostrovityanka [42]

Answer:

C. 3, -7

Step-by-step explanation:

x^2 +4x-21 =0

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What 2 numbers multiply to -21 and add to 4

7*-3 = -21

7-3 = 4

(x+7) (x-3) =0

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x+7 =0 x-3=0

x+7-7=0-7 x-3+3=0+3

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3 years ago

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21 divided by 15.015 step my step diving decimals

ser-zykov [4K]

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0.715

Step-by-step explanation:

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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

.can someonee help me as soon as possible​

.can someonee help me as soon as possible