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Debora [2.8K]
2 years ago
13

Which of the following describes an editor when writing a program?

Engineering
2 answers:
Sergeu [11.5K]2 years ago
8 0

Answer:

A place where code is written.

miskamm [114]2 years ago
6 0

Answer:

B. A person who checks for spelling errors

Explanation:

Editors typically do the following: Read content and correct spelling, punctuation, and grammatical errors. Rewrite text to make it easier for readers to understand. Verify facts cited in material for publication.

You might be interested in
A program contains the following function definition: int cube(int number) { return number * number * number; } Write a stateme
Nonamiya [84]

Answer:

The statement can be written as

int result = cube(4);

Explanation:

A function is a block of reusable codes to perform some tasks. For example, the function in the question is to calculate the cube of a number.

A function can also operate on one or more input value (argument) and return a result. The <em>cube </em>function in the question accept one input value through its parameter <em>number </em>and the <em>number</em> will be multiplied by itself twice and return the result.  

To call a function, just simply write the function name followed with parenthesis (e.g. <em>cube()</em>). Within the parenthesis, we can include zero or one or more than one values as argument(s) (e.g. <em>cube(4)</em>).

We can then use the "=" operator to assign the return output of the function to a variable (e.g. <em>int result = cube(4)</em>)

8 0
3 years ago
Air expands through a turbine operating at steady state. At the inlet p1 = 150 lbf/in^2, T1 = 1400R and at the exit p2 = 14.8 lb
Paraphin [41]

Answer:

The power developed in HP is 2702.7hp

Explanation:

Given details.

P1 = 150 lbf/in^2,

T1 = 1400°R

P2 = 14.8 lbf/in^2,

T2 = 700°R

Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h

Using air table to obtain the values for h1 and h2 at T1 and T2

h1 at T1 = 1400°R = 342.9 Btu/h

h2 at T2 = 700°R = 167.6 Btu/h

Using;

Q - W + m(h1) - m(h2) = 0

W = Q - m (h2 -h1)

W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h

W = (-65000 Btu/h ) - (-1928.3) Btu/s

W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s

W = -18.06Btu/s + 1928.3 Btu/s

W = 1910.24Btu/s

Note; Btu/s = 1.4148532hp

W = 2702.7hp

5 0
3 years ago
Show that for a linearly separable dataset, the maximum likelihood solution for the logisitic regression model is obtained by fi
KATRIN_1 [288]

Answer:

Answer for the question:

"Show that for a linearly separable dataset, the maximum likelihood solution for the logisitic regression model is obtained by finding a weight vector w whose decision boundary wx. "

is explained in the attachment.

Explanation:

8 0
3 years ago
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas
jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine (\eta), no unit, is defined by this formula:

\eta = \frac{\dot W}{\dot Q} (1)

Where:

\dot Q - Heat input, in watts.

\dot W - Power output, in watts.

If we know that \dot W = 600\,W and \eta = 0.3, then the heat input from the combustion phase is:

\eta = \frac{\dot W}{\dot Q}

\dot Q = \frac{\dot W}{\eta}

\dot Q = \frac{600\,W}{0.3}

\dot Q = 2000\,W

The heat input from the combustion phase is 2000 watts.

8 0
2 years ago
A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is 20.3 kJ/K
Mashcka [7]

Answer:

W= 8120 KJ

Explanation:

Given that

Process is isothermal ,it means that temperature of the gas will remain constant.

T₁=T₂ = 400 K

The change in the entropy given ΔS = 20.3 KJ/K

Lets take heat transfer is Q ,then entropy change can be written as

\Delta S=\dfrac{Q}{T}

Now by putting the values

20.3=\dfrac{Q}{400}

Q= 20.3 x 400 KJ

Q= 8120 KJ

The heat transfer ,Q= 8120 KJ

From first law of thermodynamics

Q = ΔU + W

ΔU =Change in the internal energy ,W=Work

Q=Heat transfer

For ideal gas ΔU  = m Cv ΔT]

At constant temperature process ,ΔT= 0

That is why ΔU  = 0

Q = ΔU + W

Q = 0+ W

Q=W= 8120 KJ

Work ,W= 8120 KJ

8 0
2 years ago
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