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2 years ago

Common factors of 16a and 20ab.

Mathematics

2 answers:

sweet [91]2 years ago

6 0

Answer:

$\large\boxed{Common\ factors:\ 1,\ 2,\ 4,\ a,\ 2a,\ 4a}$

Step-by-step explanation:

$16a=2\cdot2\cdot2\cdot2\cdot a\\Factors:\boxed1,\ \boxed2,\ \boxed4,\ 8,\ 16,\ \boxed{a},\ \boxed{2a},\ \boxed{4a},\ 8a,\ 16a\\\\20ab=2\cdot2\cdot5\cdot a\cdot b\\Factors:\boxed1,\ \boxed2,\ \boxed4,\ 5,\ 10,\ 20,\ \boxed{a},\ \boxed{2a},\ \boxed{4a},\ 5a,\ 10a,\ 20a,\\b,\ 2b,\ 4b,\ 5b,\ 10b,\ 20b,\ 2ab,\ 4ab,\ 5ab,\ 10ab,\ 20ab\\\\Common\ factors:\ 1,\ 2,\ 4,\ a,\ 2a,\ 4a$

san4es73 [151]2 years ago

5 0

Answer:

2, 4, and a

Step-by-step explanation:

Factors of 16a: 2, 4, 8, 16, a

Factors of 20ab: 2, 4, 5, 10, 20, a, b

The common factors of 16a and 20ab are 2, 4, and a.

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Lady_Fox [76]

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Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+6)\,\vec k$

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

The curl of $\vec F$ is

$\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-2)\,\vec k$

Then the line integral is equivalent to

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1\bigg((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+4)\,\vec k\bigg)\cdot\bigg(-u\,\vec\jmath+u\,\vec k\bigg)\,\mathrm du\,\mathrm dv$