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3 years ago

Write a matrix, that is a lower triangular matrix.

Engineering

1 answer:

shepuryov [24]3 years ago

6 0

Answer:

$\left[\begin{array}{ccc}10&0&0\\14&25&0\\57&18&39\end{array}\right]$

Explanation:

A lower triangular matrix is one whose elements above the main diagonal are zero meanwhile all the main diagonals elements and below are nonzero elements. This is one of the two existing types of triangular matrixes. Attached you will find a image referring more about triangular matrixes.

If there is any question, just let me know.

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Explain the LWD process why is it important in drilling operations?

barxatty [35]

Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible.

Explanation:

pls mark brainliest

6 0

1 year ago

An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formatio

saveliy_v [14]

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

$K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}$

Put the value into the formula

$K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}$

$K_{H}=41.9\ m/d$

We need to calculate the vertical conductivity

Using formula of vertical conductivity

$K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}$

Put the value into the formula

$K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}$

$K_{V}=37.2\ m/d$

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

3 0

2 years ago

A company, studying the relation between job satisfaction and length of service of employees, classified employees into three le

Wewaii [24]

Answer:

Below see details

Explanation:

A) It is attached. Please see the picture

B) First to calculate the overall mean,

μ=65∗25/75+80∗25/75+95∗25/75

μ=65∗25/75+80∗25/75+95∗25/75 = 80

Next to calculate E(MSTR) = σ2+(1/r−1) ∑ni(μi−μ)^2 = 5634

And E(MSE) = σ^2= 9

C) Yes, it is substantially large than E(MSE) in this case.

D) If we sampled 25 employees from each group, we are likely to get a F statistics to indicate differences of job satisfactions among three types of length of service of employees.

3 0

2 years ago

What is the least count of screw gauge?<br> (a) 0.01 cm<br> (b) 0.001 cm<br> (c) 0.1 cm<br> (d) 1 mm

Nonamiya [84]

Its 0.001

0.01 x100 = 1mm
0.001x100=0.1mm
0.1=10mm
1m

3 0

2 years ago

The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in

amm1812

Answer:

The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)

Explanation:

Since the given volume flow rate is gallons per minute.

We know that 1 gallon = 3.785 liters and

1 minute = 60 seconds

Let the flow rate be $Q\frac{gallons}{minute}$

Now replacing the gallon and the minute by the above values we get

$Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}$

Thus $Q'=0.631Q\frac{liters}{second}$

Now since we know that 1 liter = $0.0353ft^{3}$

Using this in above relation we get

$Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q$

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.

3 0

2 years ago