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3 years ago

the lenght of a rope is 2.589 yards round the lenght to the nearest tenth of a yard.2.589 yards rounds to_________yards.

Mathematics

1 answer:

andrezito [222]3 years ago

8 0

2.6 yards. You're welcome

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3 years ago

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How do you make a table for a function whose graph has a y-intercept of 4 and a slope of 2

ch4aika [34]

Point of Y intercept is (0,4)
Y1-Y2=m(X1-X2)
Plug in x,y into X2, Y2 m=slope

y-4=2(x-0)
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3 years ago

A hospital recorded the weight of babies born at the hospital during one week on a line plot. How many more pounds was the baby

lbvjy [14]

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2 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

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

4 0

3 years ago

1/4 kilometer = how many meters

Alchen [17]

The answer to your question is
1/4 kilometer = 0.25 kilometers = 250 meters

You answer is 250 meters

3 0

3 years ago