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Vilka [71]
4 years ago
5

Please help me! I will mark you brainliest if you are right and show your work!! I don't get polynomials and all the other stuff

.
Multiply and simplify.

2x(^2)y(^3)z(^2) · 4xy(^4)x(^2)
Show your work!!
Mathematics
2 answers:
Solnce55 [7]4 years ago
4 0
Is that x at the end supposed to be a z?
lord [1]4 years ago
4 0

\bf 2x^2y^3z^2\cdot 4xy^4x^2\implies 2\cdot 4\cdot x^2\cdot x^2\cdot x\cdot y^3\cdot y^4\cdot z^2\implies 8x^{2+2+1}y^{3+4}z^2 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 8x^5y^7z^2~\hfill

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Points A,B, and C are collinear. Points M and N are the midpoints of segments AB and AC. Prove that BC = 2MN
irinina [24]
Look at the picture.

1)|AM| = |MB| = x
|AN| = |NC| = y
|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|
2)|AM| = |MB| = x
|AN| = |NC| = y

|BC| = 2y - 2x = 2(y - x)
|MN| = y - x
Therefore |BC| = 2|MN|

6 0
4 years ago
Read 2 more answers
Calculate the flux of the vector field F⃗ (x,y,z)=(exy+9z+4)i⃗ +(exy+4z+9)j⃗ +(9z+exy)k⃗ through the square of side length 3 wit
ikadub [295]

The square (call it S) has one vertex at the origin (0, 0, 0) and one edge on the y-axis, which tells us another vertex is (0, 3, 0). The normal vector to the plane is \vec n=\vec\imath-\vec k, which is enough information to figure out the equation of the plane containing S:

(x\,\vec\imath+y\,\vec\jmath+z\,\vec k)\cdot(\vec\imath-\vec k)=0\implies x-z=0\implies z=x

We can parameterize this surface by

\vec s(x,y)=x\,\vec\imath+y\,\vec\jmath+x\,\vec k

for 0\le x\le\frac3{\sqrt2} and 0\le y\le3. Then the flux of \vec F, assumed to be

\vec F(x,y,z)=(e^{xy}+9z+4)\,\vec\imath+(e^{xy}+4z+9)\,\vec\jmath+(9ze^{xy})\,\vec k,

is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\iint_S\vec F(\vec s(x,y))\cdot\vec n\,\mathrm dx\,\mathrm dy

=\displaystyle\int_0^3\int_0^{3/\sqrt2}\left((4+e^{xy}+9x)\,\vec\imath+(9+e^{xy}+4x)\,\vec\jmath+(e^{xy}+9x)\,\vec k\right)\cdot(\vec\imath-\vec k)\,\mathrm dx\,\mathrm dy

=\displaystyle\int_0^3\int_0^{3/\sqrt2}4\,\mathrm dx\,\mathrm dy=\boxed{18\sqrt2}

3 0
3 years ago
A boat is being towed such that the tow line is at an angle of 30° with respect to the surface of the ocean. If the boat is towe
pentagon [3]

Answer:

C. 303,000 Joules

Step-by-step explanation:

1. use abcos(theta)

a=50

b=7000

theta=30 degrees

2. (50)(7000)cos(30)=303,000

5 0
3 years ago
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Can someone explain how to do this and answer 5-8
Shkiper50 [21]
#5 is 96 because it is congruent to the one on top
6 0
4 years ago
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What is (57/tan30°)m^2
iVinArrow [24]
57\sqrt{3}m^2. Enjoy!
6 0
4 years ago
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