A triangle △ABC has two sides of length x. What are all the possible types of △ABC?

G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =

Svetach [21]

Presumably you should be doing this using calculus methods, namely computing the surface integral along $\mathbf r(u,v)$ .

But since $\mathbf r(u,v)$ describes a sphere, we can simply recall that the surface area of a sphere of radius $a$ is $4\pi a^2$ .

In calculus terms, we would first find an expression for the surface element, which is given by

$\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k$
$\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j$
$\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k$
$\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u$

So the area of the surface is

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u$
$=-2\pi a^2(\cos\pi-\cos 0)$
$=-2\pi a^2(-1-1)$
$=4\pi a^2$

as expected.

6 0

2 years ago

Could someone explain how they get the answer for this?

jarptica [38.1K]

Answer:

40°

Step-by-step explanation:

The reference angle is the positive acute angle created by the terminal arm and the x-axis.

The highlighted red in the picture below shows what we're looking for.

The arm rotated 220° (but 'backwards' so the value given is negative).

|-220°| - 2(90°) <= Subtract two right angles for two quadrants

= 220° - 2(90°)

= 220° - 180°

= 40°

Therefore, the reference angle is 40°.

If you got 50°, you probably calculated the angle with the terminal arm and the y-axis. Remember to always use the nearest side of the x-axis!

4 0

3 years ago

Solve |2x - 2| < 8

Alex777 [14]

Answer:

C

Step-by-step explanation:

|2x - 2| < 8

it implies that

2x - 2 < 8

2x -2 + 2 < 8 + 2

2x < 10

x < 5

or

-(2x - 2) < 8

-2x + 2 < 8

-2x + 2 - 2 < 8 - 2

- 2x < 6

x > -3 ( the sense of the inequality reverse because I divided by minus)

The solution can be combine to

-3<x<5

4 0

2 years ago

A backyard pool has a concrete walkway around it that is 3 feet wide on all sides. The total area of the pool and the walkway is

Gnoma [55]

Answer:

<h2><em>29.4 length and 21.4 width.</em></h2>

Step-by-step explanation:

To fin the dimensions of the pool we need to use the total area of the pool plus the concrete walkway. Lets say that <em>W </em>is the width and <em>L </em>the length. We know that, the walkway has 3 ft wide, this means that the width and length are increased by 6ft (3ft on each side). So, the total area can be expressed like this: $A_{total} = (L+6)(W+6)$ .

In addition, we know that the total area is 970 and the length is 8 ft longer than the width ( $L=W+8$ ). So:

$A_{total} = (L+6)(W+6)\\970 = (W+8+6)(W+6)\\970=W^{2}+20W+84\\W^{2}+20W-886=0$

With this quadratic equation we're gonna have to solution, the most reasonable is part of the answer. So, solving the equation we have as solutions: $W_{1}=21.4$ and $W_{2}=-41.4$