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3 years ago

How many paragraphs are good for an essay

Engineering

2 answers:

zhannawk [14.2K]3 years ago

3 0

Answer:

It depends what type of essay you're writing. Normally you'll want an intro, three body paragraphs and then a conclusion, but if you're writing an argumentative essay you will want an intro par, three body par, a counterclaim and rebuttal par, and a conclusion par.

Explanation:

Typically 5 or 6 paragraphs will work, really just depends what type of essay.

NISA [10]3 years ago

3 0

Answer:

In its simplest form, an essay can consist of three paragraphs with one paragraph being devoted to each section. Proponents of the five paragraph essay say that the body text should consist of three paragraphs, but in reality, it's fine to write more or fewer paragraphs in this section. An essay is split into sections not paragraphs. While shorter essays will often have one paragraph per section most will have multiple. Paragraph and section are not interchangeable terms. If you wish to only have two paragraphs then you will need to make the core point sections into one paragraph. An essay can be pretty much any length, as long as it's relatively short. Some essays are just one paragraph, and some are several pages long. As long as you say what needs to be said, you can write four paragraphs, six paragraphs, or thirty paragraphs. 1500 words is 8 to 15 paragraphs for essays, 15 to 30 paragraphs for easy writing.

Explanation:

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Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state

lord [1]

Answer:

W=-940.36 KJ

Explanation:

Given that

$P_1=1\ bar,V_1=4.25 {m^3}$

$P_2=6.8\ bar$

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

Now by putting the values (1.4 bar =140 KPa)

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=140\times 4.25 \ln \dfrac{1.4}{6.8}$

W=-940.36 KJ

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7 0

4 years ago

In the truss below, the vertical reaction force at A is

Montano1993 [528]

It’s c trust just trust

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3 years ago

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T

anyanavicka [17]

Answer:

R min = 28.173 ohm

R max = 1.55 × $10^{4}$ ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates = 0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = $V_o \times e^{t\RC}$ ...........1

so R will be

R = $\frac{t}{C\times ln(\frac{V_o}{V})}$ ....................2

put here value

so for t min 11.5 μs

R = $\frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}$

R min = 28.173 ohm

and

for t max 6.33 ms

R max = $\frac{6.33}{11.5} \times 28.173$

R max = 1.55 × $10^{4}$ ohm

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Finger [1]

Answer:

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All except : Database server

Explanation:

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Answer:

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