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SOVA2 [1]
4 years ago
13

Mr. Vella can build a brick wall in 4 days. His apprentice can build the same wall in 6 days. After working alone for 3 days, Mr

. Vella became ill and left the job for his apprentice to complete. How many days did it take the apprentice to finish the wall?
Mathematics
1 answer:
Vitek1552 [10]4 years ago
6 0
Mr. Vella can build the wall in 4 days, however only works on it for 3 days. That means that he has completed 75% of the and is leaving 25% for his apprentice.

We need to figure out how long it takes the apprentice to build 25% of a wall. We know he can build 100% of a wall in 6 days, so dividing 6 days by 4 will give us our answer.

6/4 = 3/2 = 1.5 days

It took the apprentice 1.5 days to finish the wall.
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Given the following functions:
andrew-mc [135]

Answer:

x^2 - 6x + 9.

Step-by-step explanation:

f(x) = x^2 and g(x) = x - 3.

To find f(g(x))  we replace the x in f(x) by g(x).

f(g(x)) = (x - 3)^2

= x^2 - 6x + 9.

5 0
3 years ago
At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
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Answer:

it depends in u

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