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2 years ago

Black and white squares codehs, i need the whole code (40 points for correct answer)

Computers and Technology

2 answers:

ludmilkaskok [199]2 years ago

3 0

Answer:

speed(0)

penup()

setposition(-100,0)

count=0

def make_squares(i):

if i % 2 == 0:

begin_fill()

for i in range(4):

forward(25);

left(90)

end_fill()

penup()

pendown()

for i in range(6):

pendown()

make_squares(i)

penup()

forward(35)

Explanation:

atroni [7]2 years ago

3 0

Answer:

Answer:

speed(0)

penup()

setposition(-100,0)

count=0

def make_squares(i):

if i % 2 == 0:

begin_fill()

for i in range(4):

forward(25);

left(90)

end_fill()

penup()

pendown()

for i in range(6):

pendown()

make_squares(i)

penup()

forward(35)

Explanation:

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1 year ago

To move an icon, you right-click it and select “Pin to start menu.” True False

tester [92]

"To move an icon, you right-click it and select “Pin to start menu.” True False"

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2 years ago

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Your uncle spent most of his teen years in a hospital undergoing treatment for a severe physical illness. As an adult, he is rat

ziro4ka [17]

Answer:

one-dimensional

Explanation:

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3 years ago

Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For f

julia-pushkina [17]

Complete Question:

Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?

Answer:

a) 51.2 msec. b) 5.12 msec

Explanation:

If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:

Tw = K*512* bit time

The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:

a) BW = 1 Mbps = 10⁶ bps

⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec

b) BW = 10 Mbps = 10⁷ bps

⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec

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3 years ago

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