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beks73 [17]
2 years ago
15

How were settlement houses and political machines similar? How were they different?

Mathematics
1 answer:
Whitepunk [10]2 years ago
7 0

Answer:

OK BTW It's gonna be very long

Step-by-step explanation:

<h3><em><u>Political </u></em><em><u>machines </u></em></h3>

Political machine, in U.S politics, a party organization, headed by a single boss or small autocr atic group, that commands enough votes to maintain political and administrative control of a city, county, or state.

The rapid growth of American cities in the 19th century, a result of both immig ration and migra tion from rural areas, created huge problems for city govern ments, which were often poorly structured and unable to provide services. In those condi tions, politi cal machi nes—such as Tammany Hall, run by boss William Magear Tweed (1823–73) in New York City—were able to build a loyal voter following, especi ally among immig rant groups, by performing such favours as provi ding jobs or housing.

Political machines are charac terized by a discipli ned and hierarc hical organiza tion, reaching down to neighbourh ood and block organizers, that enables the machine to respond to the problems of individual neighbourhoo ds, or even families, in exchange for loyalty at the polls. The term refers to their ability to elect candidates or enact measures with mechanical efficiency and predictability.

<h2><em><u>The </u></em><em><u>settlement </u></em><em><u>houses </u></em></h2>

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Tan is the right answer

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Answer: The quotient is (x-2).

Step-by-step explanation:

Since we have given that

f(x)=(x^3+3x^2-4x-12)\\\\and\\\\g(x)=x^2+5x+6\\\\So,\ \frac{\left(x^3+3x^2-4x-12\right)}{\left(x^2+5x+6\right)}

Now, we have to find the quotient of the above expression.

So, here we go:

Factorise\ (x^3+3x^2-4x-12)\\\\=\left(x^3+3x^2\right)+\left(-4x-12\right)\\\\=-4\left(x+3\right)+x^2\left(x+3\right)\\\\=\left(x+3\right)\left(x^2-4\right)

Now, we will divide the above simplest form with g(x):

\frac{\left(x+3\right)\left(x^2-4\right)}{\left(x+2\right)\left(x+3\right)}\\\\=\frac{x^2-4}{x+2}\\\\=\frac{\left(x+2\right)\left(x-2\right)}{x+2}\ using\ (a^2-b^2)=(a+b)(a-b)\\\\=x-2

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Question #2:
Lesechka [4]

Answer:

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Step-by-step explanation:

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