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irga5000 [103]
3 years ago
11

What is the first eight multiples of 21

Mathematics
1 answer:
tangare [24]3 years ago
4 0
The of the first multiples of 21 is 21, 42, 63, 84, 105
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Who knows what pi(3.14)x-99.29?
MrMuchimi

Answer:

Pi also appears in the calculations to determine the area of an ellipse and in finding the radius, surface area, and volume of a sphere.  

Step-by-step explanation:

The number represented by pi (π) is used in calculations whenever something round (or nearly so) is involved, such as for circles, spheres, cylinders, cones, and ellipses. Its value is necessary to compute many important quantities about these shapes, such as understanding the relationship between a circle’s radius and its circumference and area (circumference=2πr; area=πr²).

Our world contains many round and near-round objects; finding the exact value of pi helps us build, manufacture, and work with them more accurately.

6 0
3 years ago
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What is the answer to 8x - 6x = -18
SashulF [63]
8X - 6X = -18

2X = -18

X = -9.

Hope this helps!
5 0
3 years ago
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I NEED HELP VOLUME OF SPHERES 5 QUESTIONS ASAP!!!!
Alexeev081 [22]

Answer:

1.)4188.79

2.) 7238.23

3.)1.02

4.)4189

5.)170 cm³

Step-by-step explanation:

1.) 4πr2 * 5

2.) 4πr2 * 12

3.) V = 4/3(PI*r3). *4.5

4.) V = 4/3 π r ^3 * 10

5.) Given:

Cylindrical container: height = 18 cm ; diameter = 6 cm.

3 balls each have a radius of 3 cm.

Volume of a cylinder = π r² h

V = 3.14 * (3cm)² * 18 cm

V = 508.68 cm³

Volume of rubber ball = 4/3  π r³

V = 4/3 * 3.14 * (3cm)³

V = 113.04 cm³

113.04 cm³  * 3 balls = 339.12 cm³

508.68 cm³ - 339.12 cm³ = 169.56 cm³ or 170 cm³

There is 170 cm³ free space in the container.

6 0
3 years ago
Which coordinate point(s) is a solution to the equation 3x + 2y =12?<br> (2,3),(-2,9),(3,2)
Leno4ka [110]

Answer:

points (2,3) and (-2,9)

Step-by-step explanation:

5 0
3 years ago
Evaluate c (y + 7 sin(x)) dx + (z2 + 9 cos(y)) dy + x3 dz where c is the curve r(t) = sin(t), cos(t), sin(2t) , 0 ≤ t ≤ 2π. (hin
saw5 [17]
Treat \mathcal C as the boundary of the region \mathcal S, where \mathcal S is the part of the surface z=2xy bounded by \mathcal C. We write

\displaystyle\int_{\mathcal C}(y+7\sin x)\,\mathrm dx+(z^2+9\cos y)\,\mathrm dy+x^3\,\mathrm dz=\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r

with \mathbf f=(y+7\sin x,z^2+9\cos y,x^3).

By Stoke's theorem, the line integral is equivalent to the surface integral over \mathcal S of the curl of \mathbf f. We have


\nabla\times\mathbf f=(-2z,-3x^2,-1)

so the line integral is equivalent to

\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\mathrm d\mathbf S
=\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f\cdot\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv


where \mathbf s(u,v) is a vector-valued function that parameterizes \mathcal S. In this case, we can take

\mathbf s(u,v)=(u\cos v,u\sin v,2u^2\cos v\sin v)=(u\cos v,u\sin v,u^2\sin2v)

with 0\le u\le1 and 0\le v\le2\pi. Then

\mathrm d\mathbf S=\left(\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}\right)\,\mathrm du\,\mathrm dv=(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv

and the integral becomes

\displaystyle\iint_{\mathcal S}(-2u^2\sin2v,-3u^2\cos^2v,-1)\cdot(2u^2\cos v,2u^2\sin v,-u)\,\mathrm du\,\mathrm dv
=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u-6u^4\sin^3v-4u^4\cos v\sin2v\,\mathrm du\,\mathrm dv=\pi<span />
4 0
3 years ago
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