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Y_Kistochka [10]
4 years ago
5

Y<3x+4 in graph help me please!!!!

Mathematics
1 answer:
Zinaida [17]4 years ago
7 0
Graph the boundary line of y = 3x + 4 first.  This is pretty easy. I'll give the steps below.
1. plot the y intercept:  we see from the equation, the y intercept is 4 or specifically the point (0,4).... plot this as your first point.
2. Use the slope to get another point or couple of points.  We can see from the equation that the slope is 3(the coefficient of 'x') and can be represented as a fraction  \frac{3}{1}  This fraction implies that you would move up three units of space and to the right 1 unit of space from the y intercept to get to your next point.  
3.  Draw a broken or dashed line through these two points due to the type of inequality symbol you have in the original problem(a less than symbol).  This means the points that lie on this line will not actually by solutions.
4. Now that you have the boundary line sketched, all the points falling below the boundary line will be the solutions to the inequality <span>Y<3x+4 So shade that region of your graph.
5. To prove this, you can test the point (0,0) as follows: 0</span><span><3(0)+4 
or, 0</span><4  which is a true statement, meaning the point (0,0) is a solution as well as all other points on that side of the boundary line. 

Good luck ,, and I hope this helped.
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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.

<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
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