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Shtirlitz [24]
3 years ago
12

What is the sign of the product (3)(-3)(-2)(4)? (5 points) Select one: a. Positive, because the products (3)(-3) and (-2)(4) are

negative, and the product of two negative numbers is positive b. Positive, because the products (3)(-3) and (-2)(4) are positive, and the product of two positive numbers is positive c. Negative, because the products (3)(-3) and (-2)(4) are negative, and the product of two negative numbers is negative d. Negative, because the products (3)(-3) and (-2)(4) are positive, and the product of two positive numbers is negative
Mathematics
1 answer:
Gnesinka [82]3 years ago
4 0

Answer:  a. Positive, because the products (3)(-3) and (-2)(4) are negative, and the product of two negative numbers is positive

Step-by-step explanation:  It is also possible to rearrange the factors.

(3)(4) = 12  a positive number.

(-3)(-2) = 6, a positive number. Again "a negative times a negative is a positive"  --Jaime Escalante, <em>Stand and Deliver</em>

Now you have two positive numbers: 12×6 = 72

The product is positive either way.

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Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

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In calculus terms, we would first find an expression for the surface element, which is given by

\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k
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\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u
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as expected.
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