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4 years ago

Describe how to use the formula C = StartFraction 5 Over 9 EndFraction (F minus 32) to find

Mathematics

1 answer:

MArishka [77]4 years ago

7 0

Answer:

Answer is below

Step-by-step explanation:

C = 5/9(F - 32) This equation is used to convert from Fahrenheit to Celsius on the temperature scale.

An example is listed below

C = 5/9(32 - 32) Subtract in the parentheses

C = 5/9(0) Multiply

C = 0 When it is 32 degrees Fahrenheit, it is 0 degrees Celsius

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the ratio of boys to girls in a homeroom at berkmar middle school is 3;4 if there are 12 boys how many girls are in homeroom

REY [17]

Answer:

16 girls

Step-by-step explanation:

If you have 12 boys, and the ratio is 3;4, that means you multiplied 3 times 4. To find the answer to this problem, you need to multiply 4 by 4 because you multiplied 3 by 4, and you always have to multiply or divide by the same thing if its a ratio.

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3 years ago

Help fast please! <3

Gnesinka [82]

Answer:

Step-by-step explanation:

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4 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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4 years ago

What fraction of a turn is 180 degrees

EleoNora [17]

A full turn is 360 degrees. If you put 180/360 you can simplify the answer to 1/2. 180 degrees of a turn is half of a turn.

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3 years ago

Help asap 50 points! Use DESMOS to compare the 2 functions. Which direction(s) does g(x) get translated compared to f(x)?

zhuklara [117]

Right 3 units

Hope this helps :)

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3 years ago