All categories

3 years ago

Simplify 3x + 5x + 14x

Mathematics

2 answers:

navik [9.2K]3 years ago

3 0

Simply collect all the like terms together
So..3+5+14 is 22x

Dafna11 [192]3 years ago

3 0

The answer I think is 22x I could be wrong tho

You might be interested in

What is the circles equation?

kolbaska11 [484]

Answer:

A. x²+y² = 9

Step-by-step explanation:

r = 3

So, r² = 9

3 0

4 years ago

How to find the slope of the line containing the pair of points?

Andru [333]

To find the slope of the line containing the pair of points, subtract y2 to y1 and divide x2-x1.
y2-y1
--------
x2-x1

5 0

3 years ago

Which angle is complementary to

Feliz [49]

Answer:

angle AOC is what i think it is but please dont go on my word wait to see what other people say first sorry

3 0

3 years ago

Read 2 more answers

A water tanker filled up to 2/3 of its height is moving with a uniform speed. On certain application of the great, the water in

son4ous [18]

Hi pupil Here's your answer :::

➡➡➡➡➡➡➡➡➡➡➡➡➡

The correct answer is option B. It will move upwards due to the inertia of motion.

⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅

Hope this helpsss......

8 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.

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