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3 years ago

Whats the square root of 10000

Mathematics

2 answers:

ANEK [815]3 years ago

7 0

Answer: 100

Step-by-step explanation:

Serga [27]3 years ago

3 0

Answer:

The square root of 10000 is 100

Step-by-step explanation:

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Round this number to the nearest tenth. 52.85

k0ka [10]

Answer:

52.9

numbers 0-4 the number stays the same

numbers 5-9 the number goes up one

6 0

3 years ago

Find the product. 5x2.2y2

ozzi

Answer:10x^2y^2

Step-by-step explanation:

5x2 X 2y2=10x^2y^2

3 0

3 years ago

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

Help pleaseeeeeeeeee

sergij07 [2.7K]

9514 1404 393

Answer:

47 -6√10

Step-by-step explanation:

As you know, the area of a square is the square of the side length. It can be helpful here to make use of the form for the square of a binomial.

(a -b)² = a² - 2ab + b²

(√2 -3√5)² = (√2)² - 2(√2)(3√5) + (3√5)²

= 2 - 6√10 + 3²(5)

= 47 -6√10

Check

√2-3√5 ≈ -5.29399 . . . . . . . . note that a negative value for side length makes no sense, so this isn't about geometry, it's about binomials and radicals

(√2-3√5)² ≈ 28.02633

47 -6√10 ≈ 28.02633

5 0

3 years ago

Heyy pls help! worth 20 points, and please only help if you know the answer. also no links those r kinda annoying lol

iragen [17]

Step 1: Find the slope:

$m= \dfrac{y_2-y_1}{x_2-x_1}= \dfrac{-2-7}{4-(-2)} = \dfrac{-9}{6}= -\dfrac{3}{2}$

This gives you $y=-\dfrac{3}{2}x+b$ , but we need to find b.

To find b, substitute in one (x,y) pair and it doesn't matter which one. I'll go with (4,-2):

$\begin{aligned}-2&=-\dfrac{3}{2}(4)+b\\[0.5em]-2&=-6+b\\[0.5em]4&=b\end{aligned}$

Now take that b-value and plug in into the slope-intercept form:

$y=-\dfrac{3}{2}x+4$

It's always a good idea to toss in the other x-value from the other point, to make sure it checks out.

7 0

3 years ago