A1, a1r 1, a1r 2, a1r 3, a1r 4, .... a1r n-1 is the general formula for the Sum of a what?

Someone pls help ill give you a kiss and a cookie fi you help meee

Delicious77 [7]

Answer:

cscθ = 5/4, secθ = 5/3

Step-by-step explanation:

First, let's view the drawing that I have attached, with the dot in the bottom left representing the origin. We know that cosecant and secant are the reciprocals of sin and cos respectively, and in order to find sin and cos, we must find the opposite, adjacent, and hypotenuse sides. The hypotenuse is opposite the right angle, equal to √(8²+6²) = 10

Next, we can find sin and cos of θ . sinθ = 8/10=4/5, making cscθ the reciprocal of 4/5, or 5/4

Similarly, cosθ = 6/10=3/5, and secθ = 5/3

7 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

What is this question for math 2/12+1/3

Alexxandr [17]

Answer: 1/2

Multiply both the numerator and denominator of each fraction by the number that makes its denominator equal the LCD

(2/12*1/1) + (1/3+4/4)

Complete the multiplication and the equation becomes

2/12+4/12

The two fractions now have like denominators so you can add the numerators.

2+4/12 = 6/12

This fraction can be reduced by dividing both the numerator and denominator by the Greatest Common Factor

6/12 ÷ 6/6 = 1/2

Therefore:

2/12 + 1/3 = 1/2

3 0

3 years ago

Read 2 more answers

Help with this problem

Contact [7]

Table C represents a function

This is because each x value has only one y value. For the other tables, the x values of 4, 9, and 8 have 2 different y values.

3 0

3 years ago

Choose all that applies help asap

melamori03 [73]

Answer:

idk maybe its c

Step-by-step explanation:

4 0

3 years ago