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morpeh [17]
4 years ago
11

5. I, II, III, IV, V, Conjecture:

Mathematics
1 answer:
NemiM [27]4 years ago
4 0
The conjecture is each term is the corresponding Roman numeral of the precious term
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Keith mows lawns in his neighborhood. He charges a base rate of $15 per lawn and then charges $5 per hour, h, it takes him to mo
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y * ((h * x) + 15)

Step-by-step explanation:

l (Lawns)

h = hours

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y = number of lawns

y * ((h * x) + 15)

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3 years ago
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lana66690 [7]

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a

Step-by-step explanation:

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3 years ago
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
Voting for the government to institute a price below market equilibrium, the price ceiling, is a great way to get more gas to pe
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Answer:

Step-by-step explanation:

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