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4 years ago

Translate to an equation and solve the following: The difference of p and one-sixth is two-thirds.

Mathematics

2 answers:

Slav-nsk [51]4 years ago

8 0

Answer: 5/6

Step-by-step explanation:

p - 1/6 = 2/3

+1/6 +1/6

p = 5/6

Inessa [10]4 years ago

3 0

Answer:

p = 5/6

Step-by-step explanation:

For this problem, simply take the words of the problem and convert them into an equation as such:

The difference of p and one-sixth implies --> p - 1/6

is two-thirds implies --> 2/3

So, our equation:

p - 1/6 = 2/3

Now simply solve for p, and combine the fractions.

p - 1/6 = 2/3

p - 1/6 + 1/6 = 2/3 + 1/6

p = 2/3 + 1/6

p = 4/6 + 1/6

p = 5/6

Cheers.

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3 years ago

2*log(y+5)=log 20-log 5

Ilia_Sergeevich [38]

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$\frac{2 log_{10}(y + 5) }{2} = \frac{ log_{10}(20) }{2} - \frac{ log_{10}(5) }{2}$

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3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

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3 years ago