Find 350 x 2 by rewriting 350 as 35 x 10 and using the associative property

Mathematics

2 answers:

oksano4ka [1.4K]3 years ago

3 0

Answer:d 350*2 is 700.

ValentinkaMS [17]3 years ago

3 0

350 by 2 is 700 so your answer is D

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Mrs. Lazo bought 9 1/8m. Of curtain cloth. She use 3 5/6m to make a curtain for their bedroom how many meters of cloth were not

sergeinik [125]

Answer:

5 7/24

Step-by-step explanation:

First, you need to make the denominators the same. When you multiply 6 with 4 you get 24, and if you multiply 8 with 3 you get 24. Then you have to multiply the numbers you multiplied with the denominators with the numerator. So, 5 x 4 is 20 and 3 x 1 is 3. Once you put all the numbers together it should look like this: 3 20/24 and 9 3/24. In order to find the answer, you have to subtract the numbers from each other. Which would look like this: 9 3/24 - 3 20/24. But as you can see you can't subtract 3 from 20. So you have to carry the 9. This means you have to subtract 9 from 1, and then you have 27 for the numerator, this then makes it possible to subtract from 20. So then the fractions subtracted from each other is 7 and the whole numbers subtracted from each other is 5 (because the 9 is now 8 since we subtracted one from it). Whole numbers subtracted from each other: 5. Fractions subtracted from each other: 7/24. Add it together you get 5 7/24.

Numerator - the number above the fraction, ex 3 in 3/4

Denominator - the number below the fraction, ex 4 in 3/4

5 0

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

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