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3 years ago

Producers often work to maximize their and make them as large as possible.

SAT

2 answers:

lakkis [162]3 years ago

7 0

Answer:

profits

Explanation:

yKpoI14uk [10]3 years ago

6 0

Producers often work to maximize product to maximize profit with cause and effect

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What was the effect of the civilian conservation corps in south carolina brainly

loris [4]

Answer:

In South Carolina, the CCC created 16 state parks that maintain some of the state's most stunning places. This work provided access to thousands of acres of general land to all South Carolinians. It even gave employment to the unemployed and helped many households endure difficult times.

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2 years ago

The vertices of square cdef are c(1, 1), d(3, 1), e(3,−1) and f(1,−1). Which of the following shows that its diagonals are congr

maks197457 [2]

The equation $-1 = -\frac {1}{1}$ shows that the diagonals are congruent perpendicular bisectors.

The vertices of the square are given as:

c = (1,1)
d = (3,1)
e =(3,-1)
f = (1,-1)

<h3>How to determine the congruent perpendicular bisectors.</h3>

Start by calculating the slope of diagonal ce using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m = \frac{-1-1}{3 -1}$

$m = \frac{-2}{2}$

$m_1 = -1$

Next, calculate the slope of diagonal df using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m = \frac{-1-1}{1-3}$

$m = \frac{-2}{-2}$

$m_2 = 1$

The slopes of both diagonals are:

$m_1 = -1$

$m_2 = 1$

By comparing both slopes, we have:

$m_1 = -\frac {1}{m_2}$

i.e.

$-1 = -\frac {1}{1}$

Hence, $-1 = -\frac {1}{1}$ shows that the diagonals are congruent perpendicular bisectors.

Read more about perpendicular bisectors at:

brainly.com/question/11006922

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2 years ago

what type of proverty is denifed by comparing an individuals income and expenses agianist a group average

kogti [31]

Answer:??????????

Explanation:

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3 years ago

ImaoImaoImaoImaoImaoImao

creativ13 [48]

Answer:

Explanation:

Dude what is going on here

6 0

3 years ago

Read 2 more answers

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