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galina1969 [7]
3 years ago
13

Complete the function remove_all(L, x) so that, given a list L and a target value x, it returns a copy of the list that excludes

all occurrences of x but preserves the order of the remaining elements.

Mathematics
1 answer:
tia_tia [17]3 years ago
4 0

Answer:

Code is given below along with explanation and output.

Step-by-step explanation:

we can use the method of list comprehension to achieve the given task.

In a list comprehension, we create a sub-list of elements which satisfy specific conditions. the condition i !=x makes sure that x would be excluded in the modified list.

def remove_all(L, x):

   return [i for i in L if i != x]

L = [1, 2, 3, 4, 5, 6, 4, 3, 2, 4 ]

print('Before =',L)

L = remove_all(L, 4)

print('After =',L)

Output:

Before = [1, 2, 3, 4, 5, 6, 4, 3, 2, 4]

After = [1, 2, 3, 5, 6, 3, 2]

As you can see we have successfully removed the multiple occurrence of x=4 while preserving the order of the list.

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Answer this with process and details plz thanks
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First part your answer is 8 second part is c.

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Sandy's mom is buying fabric for a prom dress. She already has 4 1/4 yards of fabric at home, and she needs a total of 17 3/6 ya
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Sandy will need 13 3/12 more fabric
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:<br> Solve the inequality and graph the solution.<br> ​<br> ​ 3(x+5)&gt;12
AleksandrR [38]

Answer:

​3(x + 5) > 12

x + 5 > 12/3

x + 5 > 4

x > 4 - 5

x > -1

Graph -  SEE ATTACHMENT

3 0
3 years ago
find the first fourth and tenth terms of the arithmetic sequence described by the given rule a (n)=5+(n-1)(1/2)
gulaghasi [49]
An=a1+d(n-1)
a1=first term
d=common differnce

easy

inptu 4 for n and 10

a4=5+(4-1)1/2
a4=5+3(1/2)
a4=5+3/2
a4=6 and 1/2
4th term is 6 and 1/2


10th term
a10=5+(10-1)1/2
a10=5+(9)1/2
a10=5+9/2
a10=9 and 1/2




4th term is 6 and 1/2
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5 0
3 years ago
A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the
Vedmedyk [2.9K]
(a) Using Newton's Law of Cooling, \dfrac{dT}{dt} = k(T - T_s), we have \dfrac{dT}{dt} = k(T - 75) where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get \dfrac{dT}{T - 75} = k dt. Integrate both sides to get \ln|T - 75| = kt + C.

Since T(0) = 185, we solve for C:
|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110
So we get \ln|T - 75| = kt + \ln 110. Use T(30) = 150 to solve for k:
\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)

So

\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\&#10;|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t}  \Rightarrow \\&#10;T = 75 \pm110e^{(1/30)\ln(15/22)t}

But choose Positive because T > 75. Temp of turkey can't go under.

T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\&#10;T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)}  = 136.929 \approx 137{}^{\circ}F

(b)

T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30}  \\&#10;100 = 75 + 110(15/22)^{t/30}   \\&#10;25 = 110(15/22)^{t/30}  &#10;\frac{25}{110} = (15/22)^{t/30}   \\&#10;\ln(25/110) / ln(15/22) = t/30 \\&#10;t = 30\ln(25/110) / ln(15/22)  \approx 116\ \mathrm{min}

Dogs of the AMS.
4 0
3 years ago
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