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3 years ago

1) What are 3 important skills for teamwork and collaboration that you learned in the text and videos of this lesson?

Computers and Technology

1 answer:

Irina18 [472]3 years ago

6 0

Answer:

1) contributing

2) accepting the agreement

3) negotiating the win-win problem-related answers for attaining the goal of your team.

Well, however, there can be many that you need to follow while working in a team or collaboration. Please check the explanation.

Explanation:

Interactive indicators comprise: -

Construction and Preserving Relationships

Give and accept criticism from aristocracies or additional team associates in command to accomplish the job.

Segment credit for decent thoughts with various others.

Recognize others cleverness, involvement, originality, and contributes

Keep the ears wide open while others speak, accept the thoughts and discover intent, opinions, as well as other member's ideas.

Spread and make better the idea of the experts or other colleagues.

Etc. Please go through your prescribed book for more details.

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____________ are designed to delete temporary files (such as deleted files still in the Recycle Bin, temporary Internet files, t

Sergeu [11.5K]

Answer:

Disk Cleanup?

Explanation: I think disk cleanup would be it. Hope this helps !!

6 0

3 years ago

Read 2 more answers

We will pass you 2 inputsan list of numbersa number, N, to look forYour job is to loop through the list and find the number spec

Montano1993 [528]

Answer:

The Python code is given below

Explanation:

# Get our input from the command line

import sys

N = int(sys.argv[2])

# Convert the list of strings into integers

numbers= []

for i in sys.argv[1].split(","):

if(i.isdigit()):

numbers.append(int(i))

# numbers now contains the list of integers

f = False

#Use for loop upto len(numbers)

for i in range(0,len(numbers)):

#Use "if" loop to check

if numbers[i] == N:

#Assign "True" to "f"

f = True

#Display "i"

print(i)

break

#Check "if" loop by assigning "false" to "f"

if f==False:

#Print "-1"

print("-1")

4 0

3 years ago

Array A is not a heap. Clearly explain why does above tree not a heap? b) Using build heap procedure discussed in the class, con

Aleksandr [31]

Answer:

Sorted Array A { } = { 1, 4, 23, 32, 34, 34, 67, 78, 89, 100 }

Explanation:

Binary tree is drawn given that the binary tree do not follow both minimum heap and maximum heap property, therefore, it is not a heap.

See attached picture.

6 0

3 years ago

What nondestructive testing method requires little or no part preparation, is used to detect surface or near-surface defects in

drek231 [11]

Answer:

Eddy current inspection

Explanation:

According to my research on nondestructive testing methods, I can say that based on the information provided within the question the method being described is called Eddy current inspection. This is a method that uses electromagnetic induction to detect and characterize surface and sub-surface flaws in conductive materials

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

5 0

3 years ago

the volume of two similar solids are 1080cm and 1715cm .if the curved surface area of the smaller cone is 840cm .fond the curved

gavmur [86]

Answer:

$A_{big} = 1143.33cm^2$

Explanation:

The given parameters are:

$V_{small} = 1080$

$V_{big} = 1715$

$C_{small} = 840$

Required

Determine the curved surface area of the big cone

The volume of a cone is:

$V = \frac{1}{3}\pi r^2h$

For the big cone:

$V_{big} = \frac{1}{3}\pi R^2H$

Where

R = radius of the big cone and H = height of the big cone

For the small cone:

$V_{small} = \frac{1}{3}\pi r^2h$

Where

r = radius of the small cone and H = height of the small cone

Because both cones are similar, then:

$\frac{H}{h} = \frac{R}{r}$

and

$\frac{V_{big}}{V_{small}} = \frac{\frac{1}{3}\pi R^2H}{\frac{1}{3}\pi r^2h}$

$\frac{V_{big}}{V_{small}} = \frac{R^2H}{r^2h}$

Substitute values for Vbig and Vsmall

$\frac{1715}{1080} = \frac{R^2H}{r^2h}$

Recall that: $\frac{H}{h} = \frac{R}{r}$

So, we have:

$\frac{1715}{1080} = \frac{R^2*R}{r^2*r}$

$\frac{1715}{1080} = \frac{R^3}{r^3}$

Take cube roots of both sides

$\sqrt[3]{\frac{1715}{1080}} = \frac{R}{r}$

Factorize

$\sqrt[3]{\frac{343*5}{216*5}} = \frac{R}{r}$

$\sqrt[3]{\frac{343}{216}} = \frac{R}{r}$

$\frac{7}{6} = \frac{R}{r}$

The curved surface area is calculated as:

$Area = \pi rl$

Where

$l = slant\ height$

For the big cone:

$A_{big} = \pi RL$

For the small cone

$A_{small} = \pi rl$

Because both cones are similar, then:

$\frac{L}{l} = \frac{R}{r}$

and

$\frac{A_{big}}{A_{small}} = \frac{\pi RL}{\pi rl}$

$\frac{A_{big}}{A_{small}} = \frac{RL}{rl}$

This gives:

$\frac{A_{big}}{A_{small}} = \frac{R}{r} * \frac{L}{l}$

Recall that:

$\frac{L}{l} = \frac{R}{r}$

So, we have:

$\frac{A_{big}}{A_{small}} = \frac{R}{r} * \frac{R}{r}$

$\frac{A_{big}}{A_{small}} = (\frac{R}{r})^2$

Make $A_{big}$ the subject

$A_{big} = (\frac{R}{r})^2 * A_{small}$

Substitute values for $\frac{R}{r}$ and $A_{small}$

$A_{big} = (\frac{7}{6})^2 * 840$

$A_{big} = \frac{49}{36} * 840$

$A_{big} = \frac{49* 840}{36}$

$A_{big} = 1143.33cm^2$

<em>Hence, the curved surface area of the big cone is 1143.33cm^2</em>

4 0

3 years ago