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3 years ago

PLsss help nobody is helping me :(

Mathematics

1 answer:

nikdorinn [45]3 years ago

8 0

7. E
8.C
9. F
10.D
11.B
12.A
13.G
I think this is right

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Find the least common multiple of 8x³y6z, 6x5z², and<br> 10yzº.

kvv77 [185]

Answer:

The Least Common Multiple (LCM) of $8x^3y6z,6x5z^2, \text{ and } 10yz^0 \text{ is } 240 x^3yz^2$

Step-by-step explanation:

<u>Definition of LCM</u>

The LCM of a, b , c is the smallest multiplier that is divisible by a, b and c

Here the three terms are :

$8x^3y6z$

$6x5z^2$

$10yz^0=10y$ since $z^0 = 1$

Factoring using prime factorization we get $8x^3y6z = 2.2.2.x^3.y.2.3.z$ ³

= $2^4\cdot \:3\cdot \:x^3\cdot \:y\cdot \:z$ (1)

Factoring $6x5z^2$ we get

$2\cdot \:3\cdot \:5\cdot \:x\cdot \:z^2$ (2)

Factoring $10yz^0$ we get

$2 \;\cdot\; 5y$ ( $z^0 = 1)$ (3)

The LCM is the multiple of each of the highest power in each factor

$2^4\;.\;3\;.\;5\;.x^2\;.\;y\;.z^2 = 240 x^3yz^2$

8 0

2 years ago

Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

kifflom [539]

Looks like we have

$\vec F(x,y,z)=z^2x\,\vec\imath+\left(\dfrac{y^3}3+\sin z\right)\,\vec\jmath+(x^2z+y^2)\,\vec k$

which has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2$

By the divergence theorem, the integral of $\vec F$ across $S$ is equal to the integral of $\nabla\cdot\vec F$ over $R$ , where $R$ is the region enclosed by $S$ . Of course, $S$ is not a closed surface, but we can make it so by closing off the hemisphere $S$ by attaching it to the disk $x^2+y^2\le1$ (call it $D$ ) so that $R$ has boundary $S\cup D$ .

Then by the divergence theorem,

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV$

Compute the integral in spherical coordinates, setting

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

so that the integral is

$\displaystyle\iiint_R(x^2+y^2+z^2)\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^1\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{2\pi}5$

The integral of $\vec F$ across $S\cup D$ is equal to the integral of $\vec F$ across $S$ plus the integral across $D$ (without outward orientation, so that

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\iint_D\vec F\cdot\mathrm d\vec S$

Parameterize $D$ by

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

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k$

Then we have

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^1u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac\pi4$

Finally,

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\frac{2\pi}5-\left(-\frac\pi4\right)=\boxed{\frac{13\pi}{20}}$

6 0

4 years ago

Whats the answer for 6×6

Tatiana [17]

The correct answer is 64 bc 6 times 6 is 64

7 0

4 years ago

Read 2 more answers

Which of the following fractions is equivalent to 2:11?

Lostsunrise [7]

Answer:

Step-by-step explanation:

Step 1:

2 : 11 Ratio

Having a ratio is the same thing as dividing so if you divide 2 by 11 you get 2/11.

Step 3

2 ÷ 11

Answer:

C. 2/11

Hope This Helps :)

8 0

3 years ago

Perpendicular bisector: put the following steps in order using what you know about constructing a perpendicular bisector.

Minchanka [31]

In order to draw a perpendicular bisector you must have ruler , compass and a pencil.

Given nothing and we have to show the steps to draw a perpendicular bisector.

Perpendicular bisector is the line which divides the line into two equal parts and makes 90° angle on other line.

If we want to make a perpendicular bisector on a line then we have to follow the following steps:

1) First draw a line with the help of ruler on which we want to make perpendicular bisector.

2) Now take a compass and a pencil.

3) Put the pencil in the compass and strech the compass equal to half or little more of the length of the line.

4) Now put the compass on one end of the line and draw an arc on the line or may be just upward.

5) Now repeat the step 4 from other end.

6) Now draw a line from the point where two arcs intersects on the line.

7) It is the perpendicular bisector of the line.

Hence to draw a perpendicular bisector we require compass, ruler and a pencil.

Learn more about perpendicular bisector at brainly.com/question/11006922

#SPJ4

3 0

2 years ago