In the liner equation shown which variable represent the output values Y= MX + b

Perform the indicated operation and simplify the result.

Finger [1]

Answer:

7/ (3a-1)

Step-by-step explanation:

3a^2 -13a +4 28+7a

------------------ * --------------------

9a^2 -6a+1 a^2 -16

Factor

(3a-1)(a-4) 7(4+a)

------------------ * --------------------

(3a-1) (3a-1) (a-4)(a+4)

Cancel like terms

1 7

------------------ * --------------------

(3a-1) 1

Leaving

7/ (3a-1)

3 0

3 years ago

Can someone help me please. Question: draw and label a picture of the following terms

frutty [35]

Answer:

Please see attached

Step-by-step explanation:

3 0

2 years ago

-3x+18=7x what could you do to isolate the variable term to one side of the equation

Alexxandr [17]

Answer:

7x=−x+24 7 x = − x + 24

Step-by-step explanation:

The equations we solved in the last section simplified nicely so that we could use the. Our strategy will involve choosing one side of the equation to be the variable side, and step by step, to isolate the variable terms on one side of the equation

7 0

2 years ago

Read 2 more answers

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago

If tan x =1/2 find sin x 2

irina1246 [14]

Answer:

t

=

26

∘

57

+

k

360

∘

Step-by-step explanation:

tan

t

=

1

2

Calculator and unit circle give 2 solutions for (0, 360) -->

t

=

26

∘

57

, and

t

=

180

+

26.57

=

206

∘

57

General answer:

t

=

26

∘

57

+

k

360

∘

6 0

3 years ago