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...in base you will have

=.........

=

=

=.......

=

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{(x.y) | }
What is a ?
A relates to .
A is a that assigns to a member of , a member of the set. The key word is "".
So if you assign say as well as to number , then you a , but a . That is the logic behind the . If you draw a vertical line and it intersects the graph of the function in , then you can see that it means you have assigned both of these points to the point where your vertical line the .
An example of this is the , and you have a .

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if varies inversely with , than you have a constant of proportionality which makes

then you need to find , the constant of proportionality

... now plug in given values and

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in the form

Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form

Start with the given equation Error: 'Illegal modulus zero at (eval 1373) line 20. '. Illegal modulus zero at (eval 1373) line 20. error_location: 13

......set
...use quadratic formula to solve for
















....so, we have two complex solutions, it means there is no real solutions and there is no

and ......set


....... is at (,)

see it on a graph:


domain:



range:

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.....first graph it as equation
find and

to find , set





to find , set





plot these points and draw a line


now shade everything to the right from a line

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the relationship between and for the following
or
given:
input output
........... or



so, you are given three points which you can plot on a Cartesian coordinate system and graph a line that contains these points
(,)= (,)
(,)= (,)...this is
(,)= (,)
now we can find the equation of a line passing through these points

Solved by pluggable solver: FIND EQUATION of straight line given 2 points

hahaWe are trying to find equation of form y=ax+b, where a is slope, and b is intercept, which passes through points (x1, y1) = (1, 1) and (x2, y2) = (0, 2). Slope a is . Intercept is found from equation , or . From that, intercept b is , or . y=(-1)x + (2) Your graph:

so,the equation of a line passing through these points is where and
now we can check if all three given points lie on this line


as you can see, all three given points do lie on this line

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1.
Probability of event that occurs:



Probability of event that does not occur:
(') =
(') =
(')=

2.
A copy machine randomly puts out 10 blank sheets per 500 copies processed.

the probability that in a run of copies, sheets of paper will be blank is
≈
3.


The Binomial distribution has the probability
function
=,,,.....,
where
For people, we expect robberies.
a)






use the source to compute

Using Poisson with


b)


Using Poisson with lambda=4


c)



d)
p( highlight(x >=3))= p( 3 through 20000)




This uses normal approximation instead of Binomial because is too large and is too small.
4.
Presuming there are of total pages with misprints, there is a % probability of discovering one on any given page.
which is %

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...use quadratic formula

..you have ,, and











solutions:




or

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...first multiply

=..add

=..eliminate parentheses , be careful with a minus sign in front of parentheses

=

=

=

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1.
answer:
a.
The sum of the probabilities of complementary events is .
P(A) + P(A') =

2.

answer is: a. %
if red,the probability a person chosen has a favorite color of red is
and if blue, the probability a person chosen has a favorite color of blue is
so, the probability a person chosen has a favorite color of red and blue is ≈ or %

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make a sketch of its cross section and draw in the height
I see a right -angled triangle with height h, so that
...where and are of the base, is , and is

given:

the difference between radius of its two circular ends is
....plug in given values





as you can see, the infamous right-angled triangle

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In a triangle the sum of angles is degrees.

if angle is percent of angle ,than
.....=>...

and if angle is percent of angle , than
....=>...=>...
so, plug it in
.....solve for



=>...


=>...



so, the smallest angle of the triangle measure

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≈

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...we take only positive root
so,

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Solved by pluggable solver: FIND EQUATION of straight line given 2 points

hahaWe are trying to find equation of form y=ax+b, where a is slope, and b is intercept, which passes through points (x1, y1) = (2, 5) and (x2, y2) = (0, -4). Slope a is . Intercept is found from equation , or . From that, intercept b is , or . y=(4.5)x + (-4) Your graph:

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Simple interest is calculated on the original principal only.

where:
= principal (original amount borrowed or loaned)
= interest rate for one period
= number of periods
so, if you have a loan of $ at % annually for years, you will pay:
$
$
$
$
$

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° - °'
°' - °'
(° - °)+('-')
(°'.....answer

so, ans could not be degree minutes because you subtracting ' from '

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given:
side length

use Pythagorean theorem to find first the length of shorter diagonal :






now we can find longer diagonal () as a hypotenuse of a right-angle triangle

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a line going through (,) that is parallel to

first solve for

.......move to the right and to the left

Solved by pluggable solver: Finding the Equation of a Line Parallel or Perpendicular to a Given Line

Since any two parallel lines have the same slope we know the slope of the unknown line is (its from the slope of which is also ). Also since the unknown line goes through (4,-1), we can find the equation by plugging in this info into the point-slope formula Point-Slope Formula: where m is the slope and (,) is the given point Plug in , , and Distribute Multiply Subtract from both sides to isolate y Combine like terms So the equation of the line that is parallel to and goes through (,) is So here are the graphs of the equations and graph of the given equation (red) and graph of the line (green) that is parallel to the given graph and goes through (,)

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if you have this:
, what is when
plug in















but, if you have this:
, what is when
plug in















as you can see, in both cases the solution happens to be same

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test is OK...post your question

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roots:
real root .....=>....... =>.......
complex root ....=>.......=>....


because you have , and because denominator cannot be equal to zero. you have

a domain:



and the range is:

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here is very good explanation for your question:

about elimination

W3C click on this link

about substitution

W3C click on this link

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first draw the picture:


next, we will need to prove this:
theorem: Diagonal of any rectangle inscribed in a circle is a diameter of the circle. (
This is essentially the converse of Thales' theorem.
Let A, B, C and D be the vertices of a rectangle inscribed in a circle and let AC and BD be the diagonals of this rectangle.
We can now focus on any one of the four triangles:
, , and
Let's take :
side of this triangle is the of the rectangle
and are two sides of the rectangle, which potentially may be of different length (but we will prove they must be the if the area of the inscribed rectangle is maximized).
The angle at is the right angle since it is one of the angles of the rectangle, which by definition has four right angles.
Hence, looking at the triangle, we can see that this is a triangle inscribed in the circle.
Therefore to prove (1), we need to show that side of any such triangle must be a diameter of circle.
Proof:
Choose any three points , and on the circle and connect these points to make a triangle .
Let's suppose that the claim is that the angle at is right angle.
We will show that if this is true then it must follow that is a of the .
Connect the center of the circle () with each of the vertices of the triangle creating the segments , and .
Let's call the angle defined by the path as and the angle defined by the path as .
Since , and are all of length (equal to the length of the radius of the circle ) then (, and ) are .
We will next want to find the angle between and (the angle at made out by segments and ) using only angles and as given.
The angle between and is equal to ° (due to the fact that is and the fact that the sum of all three internal angles in a triangle sum to °).
Similarly, the angle between and is equal to °.
Finally, since all the three angles at add up to ° (full circle), it follows that the angle between and is equal to .
However, the angle at of the original triangle is equal to (follows from the fact that and triangles are ).
This angle was claimed to be ° and therefore the angle between and is equal to °.
This means that the points and and the of the circle () are .
In other words, the of the circle is on the straight line segment , which in turn means that be a of the.
So, we can conclude that of the rectangle inscribed in a circle be a circles .

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solution set:

(,)

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a)


_________________













first graph both lines as and , find and and find intersection point
...set and find
........=>.........; is at (,)

...set and find
........=>....=>....; is at (,)
--
...set and find
........=>.........; is at (,)

...set and find
........=>....=>....; is at (,)
draw a lines, note both of them are a part of solution

now shade the part where and


b)
do same with these inequalities

....=>...=>...=>......=>......=>...

..=>....=>....=>....=>..

...=>......=>.....=>...
same way you find and and draw a lines than shade the area that belongs to a solution ; it will look like this:

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=.......write as

=.............since , we have

=......write as

=..........since , we have

=

=

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...cross multiply

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the slope-intercept form:
...solve for


......the slope-intercept form where the slope is and is