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Damm [24]
4 years ago
14

Bill can paint a small house in 3 days, Sharon can paint a small house in 4 days and Wilber can paint a small house in 5 days. H

ow long will it take for Bill, Sharon and Wilber together to paint one small house? Give your answer a fraction.
Mathematics
1 answer:
Paraphin [41]4 years ago
7 0

To solve this problem, let us say that the action of totally painting a small house is called “1 job”. Therefore the rates of each person in painting the small house are:

Bill = 1 job / 3 days = 1 / 3

Sharon = 1 job / 4 days = 1 / 4

Wilber = 1 job / 5 days = 1 / 5

 

Therefore the total rate is calculated by simply adding the rates of the three.

(1 /  3) + (1 / 4) + (1 / 5) = 1 / x

where x is the number of hours it take if all them paint together

 

Calculating for by multiplying both sides by 60:

60 [(1 /  3) + (1 / 4) + (1 / 5) = 1 / x]

(60/3) + (60/4) + (60/5) = 60 / x

20 + 15 + 12 = 60 / x

47 = 60 / x

x = 60 / 47 hours

 

 

Answer:

60/47 hours

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Lilit [14]

Problem 1

We'll use the product rule to say

h(x) = f(x)*g(x)

h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)

Then plug in x = 2 and use the table to fill in the rest

h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)

h ' (2) = f ' (2)*g(2) + f(2)*g ' (2)

h ' (2) = 2*3 + 2*4

h ' (2) = 6 + 8

h ' (2) = 14

<h3>Answer: 14</h3>

============================================================

Problem 2

Now we'll use the quotient rule

h(x) = \frac{f(x)}{g(x)}\\\\h'(x) = \frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}\\\\h'(2) = \frac{f'(2)*g(2)-f(2)*g'(2)}{(g(2))^2}\\\\h'(2) = \frac{2*3-2*4}{(3)^2}\\\\h'(2) = \frac{6-8}{9}\\\\h'(2) = -\frac{2}{9}\\\\

<h3>Answer:  -2/9</h3>

============================================================

Problem 3

Use the chain rule

h(x) = f(g(x))\\\\h'(x) = f'(g(x))*g'(x)\\\\h'(2) = f'(g(2))*g'(2)\\\\h'(2) = f'(3)*g'(2)\\\\h'(2) = 3*4\\\\h'(2) = 12\\\\

<h3>Answer:  12</h3>
5 0
3 years ago
Lol do yall know the correct one, ik the answer but i doubt its correct so what do you think
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