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Mice21 [21]
3 years ago
12

Given a square matrix [A], write a single line MATLAB command that will create a new matrix [Aug] that consists of the original

matrix [A] augmented by an identity matrix [I].

Engineering
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Answer:

Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:

Aug=[A eye(size(A))]

Above command is tested in MATLAB command window and is attached in figure below

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Experimental Design Application Production engineers wish to find the optimal process for etching circuit boards quickly. They c
Veseljchak [2.6K]

Answer:

Hello your question is incomplete attached below is the missing part and answer

options :

Effect A

Effect B

Effect C

Effect D

Effect AB

Effect AC

Effect AD

Effect BC

Effect BD

Effect CD

Answer :

A  = significant

 B  = significant

C  = Non-significant

D  = Non-significant

AB  = Non-significant

AC  = significant

AD  = Non-significant

BC  = Non-significant

 BD  = Non-significant

 CD = Non-significant

Explanation:

The dependent variable here is Time

Effect of A  = significant

Effect of B  = significant

Effect of C  = Non-significant

Effect of D  = Non-significant

Effect of AB  = Non-significant

Effect of AC  = significant

Effect of AD  = Non-significant

Effect of BC  = Non-significant

Effect of BD  = Non-significant

Effect of CD = Non-significant

8 0
3 years ago
What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?
nasty-shy [4]

Answer:

0.14\ lb/in^{3}

Explanation:

Density is defined as mass ler unit volume, expressed as

\rho=\frac {m}{v}

Where m is mass, \rho is density and v is the volume. For a sphere, volume is given as

v=\frac {4\pi r^{3}}{3}

Replacing this into the formula of density then

\rho=\frac {m}{\frac {4\pi r^{3}}{3}}

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}

4 0
3 years ago
What are some common work contexts for Licensing Examiners and Inspectors? Select four options.
Akimi4 [234]

According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:

  1. Telephone
  2. Face-to-face discussions
  3. Contact with others
  4. Importance of being exact or accurate.

O*NET is an acronym for occupational information network and it refers to a free resource center or online database that is updated from time to time with several occupational definitions, so as to help the following categories of people understand the current work situation in the United States of America:

  • Workforce development professionals
  • Students
  • Human resource (HR) managers
  • Job seekers
  • Business firms

On O*NET, work contexts are typically used to describe the physical and social elements that are common to a particular profession or occupational work. Also, the less common work contexts are listed toward the bottom while common work contexts are listed toward the top.

According to O*NET, the common work contexts for Licensing Examiners and Inspectors include:

1. Telephone

2. Face-to-face discussions

3. Contact with others

4. Importance of being exact or accurate.

Read more on work contexts here: brainly.com/question/22826220

6 0
3 years ago
Read 2 more answers
If it is desired to lay off a distance of 10,000' with a total error of no more than ± 0.30 ft. If a 100' tape is used and the
Ira Lisetskai [31]

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

3 0
3 years ago
-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63
Fudgin [204]

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0
3 years ago
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