All categories

3 years ago

What is the area of the triangular flag below?

Mathematics

1 answer:

alexandr1967 [171]3 years ago

4 0

Answer:

Here is your answer

Step-by-step explanation:

PLEASE THANK, RATE AND FOLLOW ME,

AND PLEASE MARK ME AS "BRAINLIEST" ANSWER 

HOPE IT HELPS YOU 

You might be interested in

Willa says the slope of the graph is - 3. What error did

Yuri [45]

Answer:B

Step-by-step explanation: “she counted the squares instead of using the scale” by looking at the graph the first coordinates are (4,24) and instead of doing that Willa counted the squares.

5 0

3 years ago

Simply the fraction 21wx 49xy

ddd [48]

$\frac{3w}{7y}$
this is the answer

3 0

3 years ago

(SUPER EASY) ILL GIVE BRAINLIEST TOO. just answer all the questions you know.

IrinaK [193]

Answer: x,y. 2,7, 1,

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

4 years ago

2 4/10 + 3 5/8 = what does it equal

photoshop1234 [79]

2 4/10 + 3 5/8
= 6 1/40

4 0

3 years ago

Read 2 more answers