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MAVERICK [17]
3 years ago
14

Vlad spent 20 minutes on his history homework and then completely solved x math problems that each took 2 minutes to complete wh

at is the equation that can be used to find the value of y the total time thay vlad spent on his homework and what are the constraints on the value of x and y
Mathematics
1 answer:
frozen [14]3 years ago
5 0

Answer:

y = 2x + 20

y\geq 20

x\geq 0

Step-by-step explanation:

Vlad spent 20 minutes on history. Then Vlad spent 2 minutes on each math problem for homework, this is 2x. So in total, he spent 20+2x on homework. Depending on how many problems he did, the total time y can be found. This is the equation 20+2x=y.

If Vlad did 0 math problems then he spent at least 20 minutes on homework from history. This is the constraint y\geq 20.

We also know Vlad can do many math problems but the least amount he did was 0, so the second constraint is x\geq 0.

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Read 2 more answers
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Implementating the given algorithm in python 3, the greatest common divisors of <em>(</em><em>124</em><em> </em><em>and</em><em> </em><em>244</em><em>)</em><em> </em>and <em>(</em><em>4424</em><em> </em><em>and</em><em> </em><em>2111</em><em>)</em><em> </em>are 4 and 1 respectively.

The program implementation is given below and the output of the sample run is attached.

def gcd(a, b):

<em>#initialize</em><em> </em><em>a</em><em> </em><em>function</em><em> </em><em>named</em><em> </em><em>gcd</em><em> </em><em>which</em><em> </em><em>takes</em><em> </em><em>in</em><em> </em><em>two</em><em> </em><em>parameters</em><em> </em>

if a>b:

<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>greater</em><em> </em><em>than</em><em> </em><em>b</em>

return gcd (b, a)

<em>#if</em><em> </em><em>true</em><em> </em><em>interchange</em><em> </em><em>the</em><em> </em><em>Parameters</em><em> </em><em>and</em><em> </em><em>Recall</em><em> </em><em>the</em><em> </em><em>function</em><em> </em>

elif a == 0:

return b

elif a == 1:

return 1

elif((a%2 == 0)and(b%2==0)):

<em>#even</em><em> </em><em>numbers</em><em> </em><em>leave</em><em> </em><em>no</em><em> </em><em>remainder</em><em> </em><em>when</em><em> </em><em>divided</em><em> </em><em>by</em><em> </em><em>2</em><em>,</em><em> </em><em>checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>and</em><em> </em><em>b</em><em> </em><em>are</em><em> </em><em>even</em><em> </em>

return 2 * gcd(a/2, b/2)

elif((a%2 !=0) and (b%2==0)):

<em>#checks</em><em> </em><em>if</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>odd</em><em> </em><em>and</em><em> </em><em>B</em><em> </em><em>is</em><em> </em><em>even</em><em> </em>

return gcd(a, b/2)

else :

return gcd(a, b-a)

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print(gcd(124, 244))

print()

<em>#leaves</em><em> </em><em>a</em><em> </em><em>space</em><em> </em><em>after</em><em> </em><em>the</em><em> </em><em>first</em><em> </em><em>output</em><em> </em>

print(gcd(4424, 2111))

Learn more :brainly.com/question/25506437

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