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Sphinxa [80]
2 years ago
13

STAY AWAY FROM AND REPORT mariaivanovaloveyou!!!!!!! SHE IS A BRAINLY USER AND KEEPS POSTING INAPPROPRIATE PICTURES OF HER BODY

WHICH IS SO GROSS!!!!!
Computers and Technology
1 answer:
Bond [772]2 years ago
7 0

Answer:

Yep, saw her posts. Thanks for the warning though. I reported one of her posts. It's just another example of how desperate people are and what measures they will take to get attention. She'll probably be banned soon

Explanation:

Thank you for the warning though. I hope this reaches more people.

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How do you make the "Colorize" feature in GIMP consistent if you cannot connect the different places? Brainliest if you also exp
Brums [2.3K]

Ah, that's easy. You have to select certain parts by using the Free Select tool. Use colorize. Then select the next part and use colorize, but choose the latest used PRESET color to be consistent.

To cut out parts of the image, duplicate your layer without colorization and colorize the top layer. If you have already colorized, put the duplicate as the top layer and click Base Image on the edit history for your original, which is the bottom layer. Use Free Select then click Scissor Select and press the delete button on your keyboard to remove an area, then switch from top layer to bottom layer and use colorize.

6 0
3 years ago
Read 2 more answers
A 4-bit left shift register is initially in the 0000 state, with all the flip flops storing 0s. A group of bits 1011 is serially
frutty [35]

Answer:

0010

Explanation:

Serially left shifted means that the left most bit will enter the register first. The left most bit already stored in the register will move out of the sequence. The "bold" bits mentioned below highlight these left most bits:

Initial State of the Register:

0000

Group of bits entering:

1011

<u>First Clock Cycle:</u>

0000 <em>(This bold bit will move out)</em>

1011 <em>(This bold bit will move in from right side, shifting the whole sequence one place to the left).</em>

The resulting Sequence:

0001

<u>Second Clock Cycle:</u>

0001 <em>(This bold bit will move out)</em>

1011 <em>(This bold bit will move in from right side, shifting the whole sequence one place to the left).</em>

The resulting Sequence:

0010 <em>(Final Answer)</em>

6 0
3 years ago
WAp to input the radius and print<br> the area of circle
Sergeu [11.5K]

Program:-

\tt r=float(input("Enter\:the\:value\:for\:radius\:of\:circle"))

\tt x=3.14*\:r**2

\tt print("Area\:of\:Circle=\{x\}")

<h3>Sample run:-</h3>

\tt r=3

\tt Area\:of\:the\:circle=28.26

7 0
2 years ago
Fill in the blank - A generation of social-oriented, physical games such as the _______ Wii console, emerged in the late 2000s,
disa [49]

Answer:

wild bro just need points

Explanation:

6 0
3 years ago
A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2 until it reaches its final angu
Nataly_w [17]
The first thing we are going to do is find the equation of motion:
 ωf = ωi + αt
 θ = ωi*t + 1/2αt^2
 Where:
 ωf = final angular velocity
 ωi = initial angular velocity
 α = Angular acceleration
 θ = Revolutions.
 t = time.
 We have then:
 ωf = (7200) * ((2 * pi) / 60) = 753.60 rad / s
 ωi = 0
 α = 190 rad / s2
 Clearing t:
 753.60 = 0 + 190*t
 t = 753.60 / 190
 t = 3.97 s
 Then, replacing the time:
 θ1 = 0 + (1/2) * (190) * (3.97) ^ 2
 θ1 = 1494.51 rad
 For (10-3.97) s:
 θ2 = ωf * t
 θ2 = (753.60 rad / s) * (10-3.97) s
 θ2 = 4544,208 rad
 Number of final revolutions:
 θ1 + θ2 = (1494.51 rad + 4544.208 rad) * (180 / π)
 θ1 + θ2 = 961.57 rev
 Answer:
 the disk has made 961.57 rev 10.0 s after it starts up
3 0
2 years ago
Read 2 more answers
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