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Whenever a linear inequality is involved,
one side of the line is part of the solution and is shaded,
while the other side of the line is not part of the solution and is not shaded.

The dashed line in the graph below is the graph of .
That line includes all the points with .
Those points would be part of the solution of ,
but they are not part of the solution for .
To indicate that the line is not part of the solution we draw it as a dashed line.

The solution to the inequality is all the points that have .
That means that the value of the y-coordinate of those points is less that -6, so they are below the line.
So the part below the line is shaded.

There is an easy, intuitive way to figure out which side of the line to shade.
You chose a "test point" that is not on the line, and see if it is part of the solution.
If it is you shade the side of the line that contains your test point.
If the test point is not part of the solution, you shade the other side.
The origin, (0,0), is often a good test point (It makes calculations easier).
Other easy test points are (1,0), (0,1) and (1,1).
In this case, I would chose the origin, (0,0). The origin has and
To be part of the solution it would need to have , and that is not true for (0,0).
So the origin is not part of the solution, and you shade the side of the line that does not contain the origin.

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Your idea is good, but to convince yourself, you need a drawing or two (or a very good imagination). Here are my drawings.

Let's loox at the x-y plane, and consider points in region R that at part of that cross section at .
They form segment AP, that goes from the x-axis to that green line.
Points A and P have . The radius of the circle is AP
Since P is on the line, its y-coordinate is , and that is the distance AP, the radius of the circle.
So --> or

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If two supplementary angles differ by , thein their measures, in degrees are
and
Since they are supplementary their measures add to , so

From that equation, we can find
--> --> --> --> --> -->

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Between 2 and 3 there are infinite irrational numbers (and infinite rational numbers too).

ONE IDEA:
If is between 2 and 3,
then means that , which means
So is a solution, and so are
, , and .

FANCIER IDEAS:
You probably know that is an irrational number that is between 3 and 4.
So means that .
is another irrational number between 2 and 3.

FANCIER:
You may not know it, but there is an irrational number called that is approximately 2.7183.
It is sort of like in that there is no other way to express it, and to describe it you have to tell a story. However, the story of is more complicated than the story of .

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If the function has x- intercepts at -5, -3, 0, and 6, it means that f(x) is zero for those values of x.
As a consequence, when the function is factored, it must have all of the following factors:
(x-(-5))=(x+5), which becomes zero when x=-5,
(x-(-3))=(x+3), which becomes zero when x=-3,
(x-0)=x, which becomes zero when x=0, and
(x-6), which becomes zero when x=6.
(It could have other factors too, but it must have those four factors).
So . is the only option that works.

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= marked "original" price (in $)
discounted price =
The profit would be the discounted price minus the cost,
profit =
As a percentage of the $85 cost, the profit would be
x100%=20% or
--> --> --> --> --> -->
The retailer should mark the original price as $120.
The customer would get a 15% discount, which would reduce the price by $18, to $102.
Selling the shoes at $102, the retailer would have a profit of
$102 - $85 = $17, and that $17 is 20% of the $85 cost.

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You are given the coordinates of the center and one vertex, so you can plot those point right away.
You are also given the horizontal width of ellipse, the 6 units that are the length of the minor axis.
I indicated the above information in red on the diagram below:

The vertex given is 5 units below the center, so the other vertex must be 5 units above the center, at (3,7).
Since the whole minor axis is 6 units long, each end will be 3 units from the center. That puts the ends of the minor axis at (0,2) and (6,2). Those two points are called the co-vertices.
The second vertex and the co-vertices are marked in blue in the diagram.
The next and final step is sketching a curve that looks like an ellipse and goes through the vertices and co-vertices. The ellipse is drawn in green in the diagram.

If you sketch your ellipse by hand, going through the vertices and co-vertices it should be acceptable, even though the other points will not be accurately placed.

NOTE:
If it was required, you would have to locate the foci.
The distances from the center to the vertices, co-vertices, and foci are represented by a, b, and c.
= distance to vertex (semi-major axis)
= distance to co-vertex (semi-minor axis)
= distance to focus (focal distance)
They are related by
In your case, --> --> -->
So the foci wouldbe 4 units above and below the center, at (3,6) and (3,-2).

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An is sort of a stretched .


A GARDENER STORY:
If a gardener wanted to mark/draw a circular flower bed, he/she would stick a stake in the ground, and holding that loop tight around another stake, would draw a the circle in the ground with the free stake. The fixed stake, stuck in the ground, is the center of the circle. The distance from the free stake to the fixed one is constant and is the radius of the circle.

For an ellipse, the gardener sticks two stakes in the ground and draws the ellipse the same way using a third stake. The sum of the distances between all 3 stakes is constant. The fixed stakes are the foci of the ellipse. The distance between them is constant. The distances from the fixed stakes (foci) to the other stake (a point on the ellipse) is what your teacher calls the focal radii. Those distances change as the ellipse is being drawn, but their sum is constant. To draw the ellipse in yout problem you set your stakes (foci) on the x-axis), at a distance of 9 units to either side of the origin. The length of your loop of string should be the 18 unit distance between the foci, plus the 30 units sum of the focal radii.

FACTS YOU NEED TO KNOW:
The midpoint of the segment connecting the foci is the center of the ellipse, and in this case its is (0,0).
The line containing the foci contains the major axis (the longest dimension) of the ellipse, including the vertices (the two ends of the major axis).
In this case, the line containing the foci, major axis and vertices is , the x-axis.
An ellipse centered at the origin with the major axis along the x-axis has an equation of the form
= semi-major axis and = semi-minor axis
is the horizontal distance from the center to each end (vertex) to the left and right. Those left and right ends are called the vertices.
is the vertical distance from the center up and down to the top and bottom of the ellipse. (those two points are called the co-vertices. There is also a = focal distance (distance from each focus to the center).
In your problem .
The relanships below let you calculate , , and .
= sum of the focal radii (in the problem -->

So is or
SOLUTION:

-->
means --> --> -->

NOTE:
Since it is customary to make , If the major axis was along the y-, and the left and right ends would be co-vertices. Also, since it is customary to makw , they would switch the {{a}}} and {{b}}}, with under the , but otherwise the formula would be the same).

REASONS BEHIND FACTS:
Memorizing formulas may help you pass a course. Knowing the reasoning behind the formula means you do not have to memorize it, and you will remember it when faced with the same concept in a course next year (or years from now).
OP = OQ =, OF = F'O =
F'P = F'O + OP = and FP = OP - FP =
Sum of focal radii for point P = F'P + FP = , so
Sum of focal radii for all points on the ellipse =
Sum of focal radii for point Q = F'Q + FQ = -->
For the right triangle OPQ, the Pythagorean theorem says:
or

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Two points determine a line.
We can calculate the slope of the line between 2 of those points.
From the slope and the coordinates of one of the points, we can get the equation of the line. If the coordinates of the third point satisfy that equation, the third point lies on that same line.
The slope of the line connecting (a,0) to (0,b) is

Since the y-intercept is at (0,b), b is the intercept, and we can write
as the slope-intercept form of the equation of the line connecting (a,0) to (0,b).
Substituting the x-coordinate of (3a,-2b) into the equation we can find if that point lies on the same line.
For , the point on the line has
so point (3a,-2b) lies on the same line as the other two points.

If we were not asked for the equation of the line, we could calculate the slopes for two different pairs of points. If we found the same slopes connecting two of the points with the other point, that would mean they all lie on the same line.

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Dividing by we get for quotient with for a remainder.

So

For any x which satisfies the equation ,

so

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can be written as (x^3+1)/10=4/x
Multiplying both sides of the equal sign times we get the equivalent equation
--> -->
Comparing it to we find
and

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The 2 points have y=4 and y=-4.
They are on the line .
Their coordinates are (3,-4) and (3,4).
Each one forms a right triangle with other vertices A(0,0) and (3,0)
The legs of those right triangles have lengths and , and the hypotenuse has length

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This problem was posted as problem # 749493 (2013-05-16 12:20:44) and as problem # 750070 (2013-05-18 05:10:05). Each time something is was being lost in translation, but it helped to be able to listen to the message twice.

One way to translate a graph so that point (a, f(a)) moves to the origin, point (0, 0), is to replace with and with and then solve for
When we do that to
we get





The first and second derivatives of are
and so
and
Comparing to we see that the coefficient of is indeed
and is the coefficient of
so and

How would I use all of the above to find the coordinates of the point (B, C) in the graph of
that when translated to the origin turns the function into ?

I wouldn't.

I would realize that and that
which is translated 4 units to the left and 4 units down,
and that is the translation that would bring point (B, C) = (4, 4) to (0, 0).
That looks to me like the most efficient way to the solution.

Or maybe after being told that
translated turns into and that

I would realize that must have just one inflection point, just like .
Since I know that has its inflection point at (0, 0),
I would realize that the inflection point of at must be the point translated to the origin.
Then I would know that and would only need to calculate the y-coordinate of the inflection point,
--> --> --> -->

But maybe we are supposed to use the first part and realize that with it would man that translating (4, f(4)) into the origin would transform
into
and if and the equation
transforms into

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This problem was posted as problem # 749493 (2013-05-16 12:20:44) and as problem # 750070 (2013-05-18 05:10:05). Each time something is was being lost in translation, but it helped to be able to listen to the message twice.

One way to translate a graph so that point (a, f(a)) moves to the origin, point (0, 0), is to replace with and with and then solve for
When we do that to
we get





The first and second derivatives of are
and so
and
Comparing to we see that the coefficient of is indeed
and is the coefficient of
so and

How would I use all of the above to find the coordinates of the point (B, C) in the graph of
that when translated to the origin turns the function into ?

I wouldn't.

I would realize that and that
which is translated 4 units to the left and 4 units down,
and that is the translation that would bring point (B, C) = (4, 4) to (0, 0).
That looks to me like the most efficient way to the solution.

Or maybe after being told that
translated turns into and that

I would realize that must have just one inflection point, just like .
Since I know that has its inflection point at (0, 0),
I would realize that the inflection point of at must be the point translated to the origin.
Then I would know that and would only need to calculate the y-coordinate of the inflection point,
--> --> --> -->

But maybe we are supposed to use the first part and realize that with it would man that translating (4, f(4)) into the origin would transform
into
and if and the equation
transforms into

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is close to and a little less than

makes
so and

NOTE:
For other numbers, you could get at the answer by doing a prime factorization.

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AS A SYSTEM OF EQUATIONS PROBLEM:
= number of students
= number of benches
"If 5 students are seated on each bench, 12 students are left over" translates as

"If 8 students are seated on each bench, 3 benches are left over" cound be translated as
if we are creative enough to think that the 3 empty benches could have housed extra students.
Otherwise, we could translate "If 8 students are seated on each bench, 3 benches are left over" literally as
since is the number of benches that are occupied.
Of course, the two equations are equivalent:
--> --> -->
--> --> --> --> --> -->

AS A GUESS-AND-CHECK/DIVISIBILITY PROBLEM:
Since all students fit into a number of benches with students on each bench, the number of students is a multiple of .
Sitting students per bench you fill all the benches and have 12 students left over.

You could start guessing and checking from there, but it may be time consuming. You could have 4 or more benches, and 8, 16, 24, 32, 40, ... students. You can try each number, or skip some. Making a table can help. It would be better to reduce the number of choices to try.

Guessing away without further thought:
You could imagine having 4 benches. With 8 students, you can fill 1 bench with 8 students and would have 3 benches left. But with a total of 4 benches, sitting 5 on each bench you need students to fill the benches and have 12 students left over. It must be more than 8 students.
With 5 benches, you can fill 2 benches with 8 students each (a total of 16 students), and would have 3 benches left. But 5 benches at 5 students per bench would sit 25 students, so you could not fill those 5 benches and have 12 students left over, so there are more than 5 benches and more than 16 students.
With 6 benches, you could fill 3 benches with 8 students each (24 student total), and would have 3 benches left. But 6 benches at 5 students per bench would sit 30 students, so 6 benches/24 students is still too low.
Let's skip 7 benches and go to 8 benches.
With 8 benches, you could fill 5 benches with 8 students each (40 students), and would have 3 benches left. But 8 benches at 5 students per bench would sit 40 students, with no student left over, so the numbers of benches/students at 8/40 are still too low.
Let's skip further and go to 13 benches.
With 13 benches/80 students, you could fill 10 benches with 8 students each and would have 3 benches left. But 31 benches at 5 students per bench would sit 65 students, with students left over, and that is too many.
The numbers of benches/students are less than 13/80, so we may try the next smaller guess, 12 benches/72 students.
With 12 benches/72 students, you could fill 9 benches with 8 students each and would have 3 benches left. With 12 benches, at 5 students per bench, you would sit 60 students, with students left over, and that is exactly what the problem asks for. So benches and students is the answer.

ALTERNATIVELY, you can use the information to reduce the number of guesses to try to 4 benches/32 students, 12 benches/72 students, 4 benches/32 students, 20 benches/112 students, etc.

One way:
Sitting students per bench you fill all the benches and have 12 students left.T hat means you could fill more benches with another students and have just students without a seat.
That tells you that the number of students is more than a multiple of . The multiples of 5 that are not odd rather than even (like 5,15, 15, 35, etc) would not work, because adding to those numbers will still give you and odd number for the total number of students, and the total number of students is a multiple of 8, so it must be even.
Now we know that the total number of students should be an even multiple of 5 (a multiple of 10) plus 2, and it must also be a multiple of 8.
The total number of students cannot be 12, or 22, or 42, or 52, or 62, or 82, or 92, or 102, or 122 (not multiples of 8), but we should try 32, 72, and 112.

A craftier way (maybe too creative) to get at the same point:
If I gave you 8 more students, and 4 more benches you could have everyone comfortably seated at 5 students per bench.
With the 8 added students, the number of students would be a multiple of 8 and a multiple of 5, so it would be a multiple of 40.
The number of benches would be a multiple of 8 and the number of students would be a multiple of 40, as in 8/40, 16/80, 24,120.
That means that before being given the 4 extra benches/8 extra students you had
benches/ students, or
benches/ students, or
benches/ students, or ...

At 8 students per bench,
32 students use benches, using all 4 benches with no bench left over (too few); and
112 students use benches, with benches left over (too many); and
72 students use benches, with benches left over (just right).

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and so


and so


Putting it all together:

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This diagram below shows points (2,-3) and (-8, ) circled, and the distance between them


The distance between points (2,-3) and (-8,
can be calculated as
where is the difference between the x-coordinates of the points,
and is the difference between the y-coordinates of the points.
It is not a difficult calculation, and no complicated distance formula needs to be memorized. (If your teacher disagrees, and requires that you

It is not a difficult calculation, and no complicated distance formula needs to be memorized. You do not even have to worry about the absolute values or the order of the numbers you subtract to get those differences because after you square them, it does not matter if it was and ; you get the same squared difference.

NOTE: If your teacher disagrees, and requires that you make it look complicated, you may have to write something like



After finding that distance, the problem log(x-4)=1-log(x-1) + [the distance between (2,-3) and (-8, -3+ square root of 21)]
turns into
--> --> --> -->
The equation --> --> makes me suspect some typo in the problem.

IF it had been log(x-4)=-10-log(x-1) + [the distance between (2,-3) and (-8, -3+ square root of 21)] ,
it would simplify to
--> --> --> --> --> -->
with solutions and , and verifying in the original equation we would see that
is a solution of ,
but does not work because it makes and and their logarithms would not exist.

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The function is
Its graph crosses the x-axis at the point where
--> --> -->
The x-coordinate of point P is .
The slope of the tangent at point P is the value of the derivative at that point.
y'=, so the slope of the tangent at is .
Since the line tangent at P passes through the origin, its equation must be

At point P, with , -->
Since point P is on the graph of , its y-coordinate is
So --> --> --> and P is (e,2).

Now we can find the equation of :
--> --> is ,
line perpendicular to must have a slope of .
As passes through P(e,2) its equation is
--> --> -->
So , , and
The line crosses the x-axis at the point where
--> --> --> --> -->

The area of the region bounded by the curve , the straight line , and the x-axis is shown below.


can be can be calculated as the sum of:
the area below , and above the x-axis, between and ,
plus the area below between and ,

is easier than it seems.
It's just the area of the triangle with vertices (e,0), P(e,2), and (e+e/4,0)
Its base is ; its height is , and its area is .

Since ,


So and

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For and to exist, it must be that <-->
(because logarithm exist only for positive numbers).
--> --> --> --> --> -->
Since , multiplying both sides times does not require flipping the inequality sign, so
--> --> --> --> -->
The solution is

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A dotted line going through (-3,3) and (3,1) is not part of the solution (if it were it would be drawn as a solid line to indicate that), but it is a boundary.
The slope of that line is
The equation can be written in point=slope form based on point (3,-1) as
--> --> -->

The slope-intercept form of the line (the form that starts with y=...) is

The area shaded above that line represents

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There are possible teams of 5 out of 8.
The number of teams that include your friend but not you is the number of sets of 4 players they can make from the other 6 player, which is

So out of the possible teams will include your friend, but not you.
The probability of that happening is

Are you the old Locke, or the fictional future Peter Wiggin?

WARNING:
You do not need to answer.
If you answer via the "Thank you note" feature, I will see the email address you used to register (no one else will see it). A Thank you message can be private (for me only) or public (everyone could see the message, but not the email address).

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and
so
and


so

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1. If Joseph can paint a house in 10 days, Joseph can paint of the house per day (that's his rate/speed of work).
If George can paint the house in 12 days, George can paint of the house per day.
(Joseph paints a little faster than George).
During the 5 days that both work together,
Joseph will paint of the house, and
George will paint of the house.
Between the two of them, they almost have it finished in 5 days.
The fraction of the house painted during those 5 days is

The fraction of the house left to be painted is

That is the amount of work that George could do in 1 day.
Joseph, who is a little faster and could paint finishes the work in a little less than one day and gets to go home a little earlier after finishing the job the 6th day. (or maybe he will use the time to clean up his brushes and put away all materials and equipment).

2. Let's name variables.
= Garry's age (in years)
= Badong's age (in years)
= Narva’s age (in years)
Each sentence of the information given will translate into an equation and we will end up with a system of equations to solve, finding values for , and .
If Garry is one year more than twice as old as Badong, Garry's age is
(1 year more than twice B)
If the Garry and Badong together are ten years older than Narva,

Narva is 3 years younger than Garry translates as

The 3 equations together form the system of equations

Systems of equations are solved by changing one equation at a time.
You can combine 2 equations to make a more convenient equation that can replace one of the 2 equations you combined.
You do not need to write as much as I did, but you have to keep track of what direction you are going so you do not go around in circles.
We start by substituting for
--> --> --> --> --> --> -->

3. Let's name variables.
= the first number
= the second number
= the third number
The average of three numbers is 10 translates as
-->
The second (number) is one more than twice the first translates as

The third (number) is 5 more than three times the first translates as

That gives you an easy system
Substituting into the first equation the expressions for and given in the second and third equation, we get
--> --> -->
--> --> -->

4. Let's name variables.
= liters of solution A needed
= liters of solution B needed
To get 10 liters of mixture we need to make <-->
That could be one equation in a system of equations, or we could decide to use
= liters of solution B needed
and have just one variable all along.
Since brine solution A is 5% salt, liters of it will contain an amount of salt equal to

Since brine solution B is 15% salt, liters of it will contain an amount of salt equal to

The total amount of salt in the mix will be

which must equal 12% of 10 liters or
So --> --> --> -->
So you need liters of solution A and liters of solution B.
NOTE:
Notice that I did not give you units for the amount of salt.
I did not because the problem does not deserve it.
If it made a little more sense I would have said the amounts were in kilograms, but I would have to explain it as a chemical engineering mass balance problem.
To a chemist/chemical engineer (like me) this problem is total nonsense for at least 2 reasons:
1) mixing 3 liters of one solution with 7 liters of another do not necessarily give you 10 liters, and
2) concentrations should be clearly specified as in 5% w/v, where w/v means weight (w) in volume (v), meaning 5 kilograms of salt per 100 liters of brine solution (or 5 grams in 100 milliliters). If brine solution A was specified as 5% w/w (5 kilogram salt per