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serg [7]
3 years ago
13

I urgently need the answer to number 6 with the work shown

Mathematics
2 answers:
Tatiana [17]3 years ago
5 0
First option is correct
bogdanovich [222]3 years ago
3 0

Answer:

Here you go! Work shown too.

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Maria compró una blusa con valor de $300, sólo pagó $240. ¿Cuál fue el porcentaje de descuento que le hicieron?
Nostrana [21]

300 - 240 = 60

60 = 20%

Respuesta:El descuento fue del 20%.

Espero te sirva :

5 0
2 years ago
What does change mean in math PLEASE HELP AND ANSWER RIGHT AWAY IF YOU AVAILABLE
Mazyrski [523]

Answer:

"The mean is the average of the numbers: a calculated "central" value of a set of numbers. To calculate it: add up all the numbers, then divide by how many numbers there are." -- Math is Fun


4 0
3 years ago
How to calculate distance around a pool 20 by 25 if u walk around ten times​
irina1246 [14]

Answer:

900

Step-by-step explanation:

perimeter: P = 2w + 2h

P = 2(20) + 2(25)

P = 40 + 50

P = 90

Multiply perimeter by 10:

90 × 10 = 900

3 0
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.

<h2>Why?</h2>

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!

4 0
3 years ago
If f(x)=3x+5/x, what is f(a+2)?<br><br> A. 3(f(a))+5/f(a)+2<br> B. 3a+5/a+2<br> C. 3(a+2)+5/a+2
natali 33 [55]

f(x) = 3x + 5/x

f(a + 2) = 3.(a + 2) + 5/(a + 2)

Alternative C.

5 0
3 years ago
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