What is the national study of police practice?

What difficulty did russia face in industrializing its economy?.

Alona [7]

Answer:

political instability

Explanation:

Russia faced political instability during industrialization and financial problems too

4 0

2 years ago

How are the two passages different? The essay is written only for teachers, but the article is written for high-school students.

kherson [118]

The two passages are different as the essay reveals Pat Mora’s perspective, while the article has a less personal tone.

From the complete question, it should be noted that in the essay, Pat More utilized the first-person point of view and this implies that the perspective of Pat was revealed in this case.

On the other hand, the article was written in the second-person and this shows that the time is less personal than that of the essay.

Therefore, the essay reveals Pat Mora’s perspective, while the article has a less personal tone.

Read related link on:

brainly.com/question/12399959

6 0

3 years ago

When is a light year used

Alexeev081 [22]

Answer:

The light-year is most often used when expressing distances to stars and other distances on a galactic scale, especially in non-specialist and popular science publications.

Explanation:

3 0

3 years ago

F(x, y) = e−x + y2, e−y + x2 , c consists of the arc of the curve y = cos(x) from − ???? 2 , 0 to ???? 2 , 0 and the line segmen

strojnjashka [21]

Given that y = cos(x) makes up part of the boundary of C, I suspect you mean the given points to be (-π/2, 0) and (π/2, 0).

I also assume the given vector field is

$\vec F(x,y) = \left\langle e^{-x} + y^2, e^{-y} + x^2 \right\rangle$

Since $\vec F$ has no singularities on C or in its interior, Green's theorem applies:

$\displaystyle \int_C \vec F(x,y) \cdot d\vec r = \iiint_D \frac{\partial(e^{-y}+x^2)}{\partial x} - \frac{\partial(e^{-x}+y^2)}{\partial y} \, dA = 2 \iiint_D (x + y) \, dA$

where D is the interior of C, the region

$D = \left\{ (x, y) : -\dfrac\pi2 \le x \le \dfrac\pi2 \text{ and } 0 \le y \le \cos(x) \right\}$

The integral then reduces to

$\displaystyle 2 \iint_D (x + y) \, dA = 2 \int_{-\frac\pi2}^{\frac\pi2} \int_0^{\cos(x)} (x + y) \, dy \, dx$

$\displaystyle 2 \iint_D (x + y) \, dA = 2 \int_{-\frac\pi2}^{\frac\pi2} \left( x\cos(x) + \frac12 \cos^2(x) \right) \, dx$

$\displaystyle 2 \iint_D (x + y) \, dA = 2 \int_{-\frac\pi2}^{\frac\pi2} \left( x\cos(x) + \frac{1 + \cos(2x)}4 \right) \, dx$

$\displaystyle 2 \iint_D (x + y) \, dA = \frac12 \int_{-\frac\pi2}^{\frac\pi2} \left( 4x\cos(x) + 1 + \cos(2x) \right) \, dx$

Since 4x cos(x) is an odd function over the symmetric interval [-π/2, π/2], its contribution to the integral is 0, and the remaining integral is trivial.

$\displaystyle 2 \iint_D (x + y) \, dA = \frac12 \int_{-\frac\pi2}^{\frac\pi2} \left( 1 + \cos(2x) \right) \, dx = \boxed{\frac\pi2}$

5 0

2 years ago

Trudy writes an expression to calculate the mass defect of a carbon-14 nucleus using the symbols in the table. Which expression

Oduvanchick [21]

Answer:

(6mp + 8mn) – mc – 14 is right option

Explanation:

7 0

2 years ago