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3 years ago

How many steps are there in the problem-solving process?

Engineering

2 answers:

riadik2000 [5.3K]3 years ago

5 0

Answer:

I think 7

Explanation:

Here are seven-steps for an effective problem-solving process.

1.Identify the issues. Be clear about what the problem is. ...

2.Understand everyone's interests. ...

3.List the possible solutions (options) ...

4.Evaluate the options. ...

5.Select an option or options. ...

6.Document the agreement(s). ...

7.Agree on contingencies, monitoring, and evaluation.

Vladimir79 [104]3 years ago

4 0

7 steps because you multiply then add then subtract then solved

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Given an integer k, a set C of n cities c1, . . . , cn, and the distances between these cities dij = d(ci , cj ), for 1 â¤ i &lt

Snezhnost [94]

Answer:

See explaination

Explanation:

2-Approximation Algorithm

Step 1: Choose any one city from the given set of cities C arbitrarily and put it in to a set H which is initially empty.

Step 2: For every city c in set C that is currently not present in set H compute min_distc = Minimum[ d(c, c1), d(c, c2), d(c, c3), ..... . . . . d(c, ci) ]

where c1, c2, ... ci are the cities in set H

and d(x, y) is the euclidean distance between city x and city y

Step 3: H = H ∪ {cx} where cx is the city have maximum value of min_dist over all possible cities c, computed in Step-2.

Step 4: Step-2 and Step-3 are iterated for k-1 times so that k cities are included int set H.

The set H is the required set of cities.

Example

Assume:-

C = {0, 1, 2, 3}

d(0,1) = 10, d(0,2) = 7, d(0,3) = 6, d(1,2) = 8, d(1,3) = 5, d(2,3) = 12

k = 3

Solution:-

Initially H = { }

Step-1: H = {0}

Step-2: Cities c \not\in H are {1, 2, 3}

min_dist1 = min{dist(0,1)} = min{10} = 10

min_dist2 = min{dist(0,2)} = min{7} = 7

min_dist3 = min{dist(0,3)} = min{6} = 6

Step-3: Max{10, 7, 6} = 10

Step-4: cx = 1

Step-5: H = H ∪ cx = {0} \cup {1} = {0, 1}

Step-6: Cities c \not\in H are {2, 3}

min_dist2 = min{dist(0,2), dist(1,2)} = min{7, 8} = 7

min_dist3 = min{dist(0,3), dist(1,3)} = min{6, 5} = 5

Step-7: Max{7, 5} = 7

Step-8: cx = 2

Step-9: H = H \cup cx = {0, 1} \cup {2} = {0, 1, 2}

Result: The set H is {0, 1, 2}.

6 0

3 years ago

A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end

hjlf

Answer:

a) $v=19.6 m/s$

b) $H=19.58 m$

c) $v_{f}=29.57 m/s$

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

$W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J$

But we know the work is equal to change of kinetic energy, so:

$W=\Delta K=\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2W}{m}}=19.6 m/s$

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

$v_{f}^{2}=v_{i}^{2}-2gh$

At the maximum height the velocity is 0, so v(f) = 0.

$0=v_{i}^{2}-2gH$

$H=\frac{19.6^{2}}{2*9.81}=19.58 m$

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

$v_{f}^{2}=v_{i}^{2}-2gh$

$v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s$

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0

3 years ago

On Beverly's last project, the team identified only a few lessons learned. Which approach to lessons learned

Afina-wow [57]

Option C (making lessons learned a regular part of meetings) is the correct approach.

As nothing more than a general rule, typically construction companies only plan lessons that have been learned exercises or initiatives towards the end of a particular endeavor or segment.
As almost a result of the team knowing, valuable lessons are intended to increase the comprehensive implementation of quality management practices as well as deadlines.

Aside from this, none of the choices are viable methods to learning lessons or gaining knowledge from the past. As a result, the methodology outlined above is the appropriate one.

Learn more about project teamwork here:

brainly.com/question/14279121

7 0

3 years ago

How can you kill someone but not kill some one but end up killing them?

sattari [20]

dkrktiroro49r9494949rototototofklfkfkrororor

5 0

3 years ago

Read 2 more answers

A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turn

Ostrovityanka [42]

Answer: a) 150 rev. b) 2105 rev.

Explanation:

a) Assuming a uniformly accelerated motion, we can use the equivalent kinematic equations, replacing linear variables by angular ones.

In order to get the number of revolutions executed, we can use this:

ωf² - ω₀² = 2 γ Δθ (1)

For the first part, we know that ω₀ = 0 (as it starts from rest).

We can find out the value of angular acceleration γ, just applying the definition of angular acceleration, as the change in angular velocity, regarding time, as follows:

γ = (ωf - ω₀) / Δt (2)

As we would want to use SI units, it is advisable to convert the value of ωf, from rpm to rad/sec.

3600 rev/min . (1min/60 sec) . (2π rad/rev) = 120π rad/sec

Replacing in (2), we get γ:

γ = 120 π / 5 rad/sec² = 24 π rad/sec²

Replacing in (1) and solving for Δθ:

Δθ = 120² π² / 2. 24 π = 300 π rad

As 1 rev = 2π rad, Δθ = 150 rev

b) For the second part, we can use exactly the same equations, taking into account that ω₀ = 120 π rad/sec, and that ωf = 0.

The new value for γ is as follows:

γ = -120π / 70 rad/sec² = -1.71 rad/sec²

Replacing in (1) and solving for Δθ, we get:

Δθ = -120² π² / 2. (-1.71) π = 4210 π rad

As 1 rev = 2π rad, Δθ = 2105 rev

7 0

4 years ago