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3 years ago

Need help with a math question PLEASE HELP

Mathematics

2 answers:

zhuklara [117]3 years ago

6 0

If you're allowed to repeat letters, then it's 12x12x12x12x12. If not, then it's 12x11x10x9x8.

julsineya [31]3 years ago

6 0

For this case we have the following formula:

$nP_ {r} = \frac {n!} {(n-r)!}$

Where:

P: It is the number of permutations

n: It is the number of objects

r: It is the number of selected objects

According to the data we have to:

$n = 12\\r = 5$

Substituting:

$12P_ {5} = \frac {12!} {(12-5)!} = \frac {12 * 11 * 10 * 9 * 8 * 7!} {7!} = 12 * 11 * 10 * 9 * 8 = 95,040$

Finally, the number of permutations is 95,040

ANswer:

95,040

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recall that the difference between a country's exports and imports is called the country's trade balance. suppose last year the

vovangra [49]

Let's call trade balance b, exports e and imports i.

We now construct formula.

$b=\Delta{ei}=e-i$

Put in the data from last year.

$b=968-1658=\boxed{-690}$

U.S trade balance for that year is -690 billion dollars.

7 0

3 years ago

Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .

Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of $F$ around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use <em>the Green's Theorem </em>to evaluate this integral. The rectangle is presented below.

We have that

$F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle$

Therefore,

$P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y$

Let's calculate the needed partial derivatives.

$P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0$

Thus,

$Q_x -P_y = 0 -2 = - 2$

Now, by the Green's theorem, we have

$\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x \Big|_{-3}^{4} = -42$