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2 years ago

4. Find the measure of x. Then find the measure of DE. Plzzz show work!

Mathematics

1 answer:

LekaFEV [45]2 years ago

6 0

Answer:

more information on how much of that they have been doing a

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Cube root -512 Simplify.

Lerok [7]

∛-512 = -8 . . . . . . . . . . matches the 3rd selection

8 0

3 years ago

A bag with 2 1/2 quarts of peanuts can make 2 1/6 jars of peanut butter. How many

diamong [38]

Answer:

You will need 3 6/13 quarts of peanuts to make 3 jars of peanuts butter

Step-by-step explanation:

To make this answering seamless, we make the mixed fractions into improper fractions

Hence 2 1/2 becomes 5/2

while 2 1/6 becomes 13/6

Here we are told that a bag with 5/2 quarts of peanuts can make 13/6 jars of peanuts butter

Hence, x quarts of peanuts will make 3 jars of peanuts butter

To get the value of x, what we simply need to do is to cross multiply.

Thus, we have

5/2 - 13/6

x - 3

3 * 5/2 = 13/6 * x

15/2 = 13x/6

26x = 90

x = 90/26

x = 45/13

x = 3 6/13

3 0

3 years ago

Answer fast again important

gtnhenbr [62]

Answer: 17 square inches.

Step-by-step explanation: Area of a triangle is 1/2 x b x h. We can just take the area of half of the whole triangle, and double it. In this case, half of the triangle has an area of 1/2 x 3.4 x 5 = 8.5. Doubling it is the same is doing 8.5 x 2, which equals 17.

5 0

2 years ago

Find the measure of each angle.

Anton [14]

Answer:

Step-by-step explanation:

1. 51 degrees

2. 90 degrees

3. 17 degrees

4. 107 degrees

5. 121 degrees

6. 68 degrees

5 0

2 years ago

Read 2 more answers

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

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago