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3 years ago

Solve the inequalities by graphing. Identify the graph that shows the following equations.

Mathematics

1 answer:

Yuki888 [10]3 years ago

3 0

Y>2x (0,0)—> (0+2, 0+1)
Y<3 Y=3 then shade downward
Horizontally

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Tomtit [17]

Apparently my answer was unclear the first time?

The flux of F across S is given by the surface integral,

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S$

Parameterize S by the vector-valued function r(u, v) defined by

$\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2. Then the surface element is

dS = n • dS

where n is the normal vector to the surface. Take it to be

$\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}$

The surface element reduces to

$\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv$

so that it points toward the origin at any point on S.

Then the integral with respect to u and v is

$\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS$

$=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS$

$=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}$

4 0

3 years ago

Need help please!!!!

ArbitrLikvidat [17]

Given : A inequality is given to us . The inequality is 19 ≥ t + 18 ≥ 11 .

To Find : The correct option between the given ones . To write the compound inequality with integers .

Solution : The given inequality to us is 19 ≥ t + 18 ≥ 11 . Let's simplify them seperately .

$\red{\tt Case\:1}$

⇒ 19 ≥ t + 18 .

⇒ t + 18 ≤ 19 .

⇒ t ≤ 19 - 18 .

⇒ t ≤ 1 = 1 ≥ t . ..................(i)

$\red{\tt Case\:2}$

⇒ t + 18 ≥ 11 .

⇒ t ≥ 11 - 18.

⇒ t ≥ -7 . ....................(ii)

On combing (i) & (ii) .

$\Large\boxed{\green{\bf \red{\longmapsto} 1 \geqslant t \geqslant -7 }}$

This means that t is less than or equal to 1 but greater than or equal to (-7) .

4 0

3 years ago

Please find angle q

mars1129 [50]

Hello!

75 + 90 + 79 + q = 360
q = 360 - 244 = 116

Stay safe!

4 0

3 years ago

The sum of all but one interior angle of a heptagon is 776°. what is the measure of the final interior angle? 42º 56º 124º 304º

Mekhanik [1.2K]

A heptagon has seven sides. The measure of the final interior angle of the heptagon is 124°.

<h3>What is heptagon?</h3>

A heptagon, sometimes known as a septagon, is a seven-sided polygon. The heptagon is also known as the septagon, which combines the prefix "sept-" with the Greek suffix "-agon," which means "angle."

A heptagon has 7 sides therefore, the sum of the interior angle of the heptagon can be written as,

$\text{Sum of intetrior angles of a polygon with n sides} = (n-2)\times 180^o\\\\\text{Sum of all the angle of heptagon} = (7-2) \times 180^o = 900^o$

As it is said in the problem that the sum of the all the angles of the heptagon except one is 776°, therefore, the measure of the last angle of the heptagon is,

The measure of the last angle = 900° - 776° = 124°

Hence, the measure of the final interior angle of the heptagon is 124°.

Learn more about Heptagon:

brainly.com/question/12630702

3 0

2 years ago

Read 2 more answers

What is 0.6n = $2.10

kompoz [17]

Answer: 3.5

Step-by-step explanation:

2.10/0.6=3.5