Y=x+3 y=3x+1 A. (1,3) B. (0,1) O C. (0,3) D. (1.4)

Mathematics

2 answers:

guajiro [1.7K]3 years ago

6 0

Answer:

D: (1, 4)

Step-by-step explanation:

You can either solve by substitution or elimination. If you are timed, the fastest way is to guess and check (plug-in the answer choices)

I will be using substitution to solve this:

Step 1: Since you have y already, you can set them equal to each other

x+3 = 3x+1

3 = 2x +1

2 = 2x

x = 1

Step 2: Now you have x, plug x into one of the original equations. In this case, I will use the first one

y = 1 + 3

y = 4

Step 3: Now write your answer as a coordinate plane point

x = 1, y = 4

(1, 4)

Murrr4er [49]3 years ago

4 0

I think it’s D hopefully I’m right

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Evaluate the surface integral. s (x + y + z) ds, s is the parallelogram with parametric equations x = u + v, y = u − v, z = 1 +

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$S$ is given to be parameterized by

$\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\langle u+v,u-v,1+2u+v\rangle$

with $0\le u\le3$ and $0\le v\le2$ . We have

$\mathbf r_u=\langle1,1,2\rangle$
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$\mathbf r_u\times\mathbf r_v=\langle3,1,-2\rangle$
$\left\|\mathbf r_u\times\mathbf r_v\right\|=\sqrt{14}$

The surface integral is then

$\displaystyle\iint_S(x+y+z)\,\mathrm dS=\iint_S(x(u,v)+y(u,v)+z(u,v))\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm du\,\mathrm dv$
$=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}((u+v)+(u-v)+(1+2u+v))\,\mathrm dv\,\mathrm du$
$=\displaystyle\sqrt{14}\int_{u=0}^{u=3}\int_{v=0}^{v=2}(4u+v+1)\,\mathrm dv\,\mathrm du$
$=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)$
$=\displaystyle\sqrt{14}\left(8\int_{u=0}^{u=3}u\,\mathrm du+3\int_{v=0}^{v=2}v\,\mathrm dv+6\right)$
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