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2 years ago

Please help ;p with math math hard last question plz help i on have 3 minutes

Mathematics

2 answers:

nika2105 [10]2 years ago

8 0

Answer:

3rd box

Step-by-step explanation:

BRAINLIEST PLEASE

Ipatiy [6.2K]2 years ago

8 0

Answer:

2(2x+20)

Step-by-step explanation:

The formula used to find the area of a rectangle is A = l * w. The formula used to find the perimeter of a rectangle is P = 2 * (l + w) In the above formulas, A stands for “area,” P stands for “perimeter,” l refers to the length of the rectangle, and w refers to the width of the rectangle.

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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:

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