Approximately what percent of the rectangle below is shaded? 25% 33% 40% 66%

Mathematics

1 answer:

trasher [3.6K]2 years ago

4 0

It’s 33 4 of 12 equals 33.333

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1)Choose one of the factors of 24x6 - 1029y3 (24x to the sixth power minus 1029y to the third power) A)3 B)2x2 - 7y C)4x4 + 14x2

grigory [225]

24x⁶ - 1029y³ =
3(8x⁶ - 343y³) =
3((2x²)³ - (7y)³) =
3(2x²-7y)(4x⁴ + 14x²y + 49y²)

Answer is <span>D) All of the above.</span>

5 0

2 years ago

Read 2 more answers

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

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