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Greeley [361]
3 years ago
11

B. Is the “Loading Time” of any online application a functional or a non-functional requirement? Can the requirement engineers s

pecify this property before the system is actually implemented, how?
Engineering
1 answer:
Oksi-84 [34.3K]3 years ago
4 0

Answer:

non-functional requirement,

Yes they can.

The application loading time is determined by testing system under various scenarios

Explanation:

non-functional requirement are requirements needed to justify application behavior.

functional requirements are requirements needed to justify what the application will do.

The loading time can be stated with some accuracy level after testing the system.

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KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown
kaheart [24]

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

i_x+50mA=i_y\\i_x-i_y=0.05A

Also

V_{\Delta}=1K*i_y

Now applying KVL on loop 1 as indicated in the attached figure

1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0

Similarly for loop 2

2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0

So the system of equations become

i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

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2 years ago
Two added to four times a number, minus 3 times the number, equals 5.
vladimir1956 [14]
<h2>Answer:</h2>

<u>x= 3</u>.

<h2>Explanation:</h2>

<em>What is presented in this problem is basically an equation in verbal form.</em>

<em />

<h3>1. Write the equation.</h3>

2+4x-3x=5

<h3>2. Solve for x.</h3>

2+4x-3x=5\\ \\2+x=5\\ \\x=5-2\\ \\x=3

<h3>3. Express the result.</h3>

x= 3.

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2 years ago
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