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4 years ago

Which was more significant to Gilded

Advanced Placement (AP)

1 answer:

fenix001 [56]4 years ago

5 0

Answer:

The answer is: the power of big businesses.

Explanation:

The Gilded Age (late 19th century) was an "age of transformation," a change in America's economy, social customs and government. The growth of the country's economy was rapid and cities expanded. Because of technological advancements such as the telephone, typewriter and the light bulb, the industrial sector became more productive. This attracted many businessmen who were eager to gain more profit, which resulted into a situation of monopoly.

These businessmen could gain an easy access to the Senate, who were responsible for implementing the nation's laws. They were favored by the Senate by giving them a favorable tariff and a businessman-friendly policy. Thus, it made the monopolists even stronger. This shows how the big businesses played a role in terms of politics and industrialization.

This is the reason why the power of the big businesses was more significant to the Gilded Age.

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Please show steps and explain

AlexFokin [52]

Answer:

$\frac{n-6}{3n} = \frac{n-5}{3n+1} \\\\\frac{n-6}{3n} - \frac{n-5}{3n+1} = 0\\\\\frac{(3n+1)(n-6)-3n(n-5)}{3n(3n+1)} = 0\\\\(3n+1)(n-6)-3n(n-5) = 0\\\\3n^2-18n+n-6-3n^2+15n=0\\\\3n^2-3n^2-18n+n+15n-6=0\\\\-2n-6=0\\\\2n+6=0\\\\2n=-6\\\\n=-3$

You can solve this equation by applying the zero product rule.

Firstly, you equate the expression to 0 in order to combine both fractions into one. As shown in Step 2, you find the LCM of both fractions and multiply accordingly. In this case, the LCM of both fractions is 3n(3n+1). Hence you cross multiply.

Next, you eliminate the denominator in order to solve the equation more conveniently. As you multiply the fraction by the denominator on the LHS, you do that to the RHS as well to maintain balance. 0×3n(3n+1)=0, hence Step 3.

Then, you expand the parentheses using the distributive law of multiplication, resulting in a quadratic equation.

Lastly, you solve the quadratic equation.

The answer is n=-3

6 0

3 years ago

Two types of monocular depth cues are

faust18 [17]

The answer is c.

I hope this helps.

5 0

3 years ago

Let S(x) = (2x + 1)^3 and let g be the inverse function of f. Given that f(0) = 1, what is the value of gʻ(1) ?

aksik [14]

Answer:

$g(1) = 0$

Explanation:

Given

$f(x) = (2x + 1)^3$

$f(0) = 1$

Required

Find g(1)

If f(x) and g(x) are inverse functions, then:

$f(x) = y$

implies that:

$g(y) = x$

Using the analysis above:

$f(0) = 1$ implies that:

$g(1) = 0$

3 0

3 years ago

(statistics) Using data from the LPGA tour, a regression analysis was performed using x = average driving distance and y = scori

inessss [21]

Answer:

(b)

A dealer gains 16% by selling a mop for Rs 78,300. But, due to competition in the market, he decides to make a profit of only 10%. What is its new selling price?

Explanation:

(b)

A dealer gains 16% by selling a mop for Rs 78,300. But, due to competition in the market, he decides to make a profit of only 10%. What is its new selling price?(b)

A dealer gains 16% by selling a mop for Rs 78,300. But, due to competition in the market, he decides to make a profit of only 10%. What is its new selling price?

7 0

1 year ago

Slope integrals for calculus, having trouble need all the help I can get

Harlamova29_29 [7]

(1) The integral is straightforward; x ranges between two constants, and y ranges between two functions of x that don't intersect.

$\displaystyle\int_{-2}^1\int_{-x}^{x^2+2}\mathrm dy\,\mathrm dx$

(2) First find where the two curves intersect:

y ² - 4 = -3y

y ² + 3y - 4 = 0

(y + 4) (y - 1) = 0

y = -4, y = 1 → x = 12, x = -3

That is, they intersect at the points (-3, 1) and (12, -4). Since x ranges between two explicit functions of y, you can capture the area with one integral if you integrate with respect to x first:

$\displaystyle\int_{-4}^1\int_{y^2-4}^{-3y}\mathrm dx\,\mathrm dy$

(3) No special tricks here, x is again bounded between two constants and y between two explicit functions of x.

$\displaystyle\int_1^5\int_0^{\frac1{x^2}}\mathrm dy\,\mathrm dx$

8 0

3 years ago