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3 years ago

A U.S. balance of trade surplus can result from

1 answer:

8 0

From the research that I have done, exports to the United States increases the country's balance of trade. Possibly creating a surplus of goods.

The correct answer would be an increase in exports to the United States

Here is a good example of what you are trying to understand.

If a country exports a greater value than it imports, it has a trade surplus, positive balance, or a "favorable balance", and conversely, if a country imports a greater value than it exports, it has a trade deficit, negative balance, "unfavorable balance", or, informally, a "trade gap". A positive balance adds to gross domestic product, while a negative balance subtracts from GDP.

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Select the correct answer.

dalvyx [7]

I would say it’s emotional support because he is helping him or supporting him when he needs help with his emotions

3 0

2 years ago

Help please its Algebra 1 honors

marissa [1.9K]

Answer:

f(x)= 50+4.25x

g(x)= 40+6.25x

(5, 71.25)

Explanation:

they are asking for the number of garmets and price of those garmets where both shops charge the same amount

if you make the two shops prices into equations you will be trying to find the line intercept

to make the lines equations

they said f(x) is cleaning company 1

g(x) is cleaning company 2

and x in the number of garmets cleaned

you can see from prices that the price does not start at 0 it has a baseline price to clean

that baseline price is the +____

113.75-92.50 to find the difference in price for 5 garmets

=21.25

21.25/5 to find raise in price per 1 garmet

=4.25

now we know its 4.25x+b

baseline price = b

to find b plug in the values on the chart (first row)

4.25(5)+b=71.25

21.25+b=71.25

b=50

baseline price = 50

store 1 price is= f(x)= 50+4.25x

im gonna run out of space ill do the second one short

103.75-72.5=31.25

31.25/5=6.25

71.25=6.25(5)+x

71.25-31.25=x

x= 40= base price

g(x)= 40+6.25x

find intersection

g(x) =y

f(x) = y

y=y

therefore

50+4.25x=40+6.25x

10+4.25x=6.25x

10=2x

x=5

to find y intersection put in x to equation

y= 50+4.25x

y= 50+4.25(5)

y=50+21.25

y=71.25

intersection is (5, 71.25)

8 0

3 years ago

The Assessment Test asked, "Do I feel disappointed when my achievements go unnoticed?" Do I Agree or Disagree?

Romashka [77]

I would disagree.
It is not always important to be noticed, because you don't have to depend on other people's approval. If you are satisfied with your accomplishments, why does it matter if someone else is going to praise you for them or not? You should be happy, regardless of other people's opinions and thoughts.

6 0

3 years ago

PLEASE ANSWER FAST

Rama09 [41]

Answer

Article 1, Section 9, Clause 8

Explanation:

I just googled it for you

7 0

3 years ago

The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded

Alex Ar [27]

Answer:

The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

y = sin²x

∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

$\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx$

Therefore;

$A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}$

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, $A_{1 shaded}$ = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

$A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526$

The shaded area, $A_{2 shaded}$ = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = $A_{1 shaded}$ + $A_{2 shaded}$

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

8 0

3 years ago