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3 years ago

Find the sum of the first 20 terms of the arithmetic series 0.5 + 1.1

Mathematics

2 answers:

exis [7]3 years ago

8 0

Answer:

0.5+1.1+1.7+2.3+2.9+3.5+4.1+4.7+5.3+5.9+6.5+7.1+7.7+8.3+8.9+9.5+10.1+10.7+11.3+11.9

Step-by-step explanation:

tell me if this isn't 20 terms

trapecia [35]3 years ago

7 0

Answer:

$S_{20}$ = 124

Step-by-step explanation:

The sum to n terms of an arithmetic series is

$S_{n}$ = $\frac{n}{2}$ [ 2a₁ + (n - 1)d ]

where a₁ is the first term and d the common difference

Here a₁ = 0.5 and d = a₂ - a₁ = 1.1 - 0.5 = 0.6 , then

$S_{20}$ = $\frac{20}{2}$ [ (2 × 0.5) + (19 × 0.6) ]

= 10 (1 + 11.4 )

= 10 × 12.4

= 124

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-15x+8y=-30 -5x+5y=-10

Scilla [17]

Answer:(x,y)=(6,10)

Step-by-step explanation:

4 0

3 years ago

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

7 0

3 years ago

The Hickory Stick has a selection of 3 meats and 6 vegetables. How many different selections of one meat and one vegetable are p

valentina_108 [34]

The different selections of one meat and one vegetable are possible are 18 selections.

Since the Hickory Stick has a selection of 3 meats and 6 vegetables, the number of ways we can select one meat out of 3 is ³C₁ = 3.

Also, the number of ways we can select one vegetable out of 6 is ⁶C₁ = 6.

So, the total number of selections of one meat and one vegetable is ³C₁ × ⁶C₁ = 3 × 6

= 18 selections

So, the different selections of one meat and one vegetable are possible are 18 selections.

Learn more about combination here:

brainly.com/question/19341024