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         x³ + 2x² - 25x - 50 = 0

Factor the first two terms on the left side by taking out GCF x²

        x²(x + 2) - 25x - 50 = 0

Now factor the last two terms on the left side by taking out GCF -25

       x²(x + 2) - 25(x + 2) = 0

Now factor out the GCF (x + 2) leaving x² and -25 in parentheses:

            (x + 2)(x² - 25) = 0

Factor the second parentheses as the difference of squares:

       (x + 2)(x - 5)(x + 5) = 0

Use the zero-factor principle by setting each factor = 0:

x + 2 = 0;  x - 5 = 0;  x + 5 =  0
    x =-2;      x = 5;      x = -5

The real-number solutions are -5, -2, and 5.

Edwin

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What is your quadratic equation? You didn't post it, so we can't tell you anything about it until we know what your quadratic equation is. Post again telling us what your quadratic equation is. Then maybe we can help you.
Edwin

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All your inequality symbols ≦ and ≧ came out as square boxes
like this ↏. So we can't help you. Next time post only 1 problem and
use symbols for ≦ and for ≧.
Or just use <= for "less than or equal to" and >= for "greater than or equal to.

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How many years will it take $6,000 to amount to $8,000 if it is invested at an annual rate of 9.0% compounded continuously?

A = Pert

A = final amount = 8000
P = beginning amount = 6000
r = rate expressed as a decimal = .09
t = the number of years = (unknown)

8000 = (6000)e(.09)t

Divide both sides by 6000

 = e.09t

 = e.09t

Since A = eB is equivalent to B = ln(A),

,09t = ln()

   t = ()÷.09

   t = 14.81481481...

Answer: about 15 years.

Edwon

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Watch these youtube videos:
http://www.youtube.com/watch?v=cuNpXve18Pc&feature=related
http://www.youtube.com/watch?v=sS0BVEvVmIQ

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I will assume you are implying that the student will not leave 
any questions blank, and will therefore either gain or lose
some points for each of the N questions.

let:

N = the number of questions.
a = the number of points added for each correct answer.
b = the number of points subtracted for each incorrect answer.
p = the probability of answering any one question correctly.
C = the number of answers a student guesses correctly.
(Therefore the student guesses N-C answers incorrectly, since
I am assuming he or she leaves no answers blank.]
X = the student's score

Then the student's highest possible score is aN and the lowest
is -bN

The formula for the score X is

X = aC - b(N - C)

Solve that for C

X = aC - bN + bC

X + bN = aC + bC

X + bN = C(a + b)

 = C

C = 

If that does not come out to a whole number, then the student
cannot possibly score exactly X, so you will then have to 
choose whether to round C up to the next higher integer so that
he or she makes slightly higher than C or to round C down to the
next lower integer and makes slightly lower than X.

The probability of scoring X (or as close to X as possible) 
is the probability of guessing C correct answers correctly,
which is:



where  is the number of combinations
of N things taken C at a time.

Edwin

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Watch these Youtube videos:
`
http://www.youtube.com/watch?v=XTZl7Kn6u4Y&feature=relmfu
`
http://www.youtube.com/watch?v=d3xr5a4cln0
`
http://www.youtube.com/watch?v=omv7Di2o8-Y
`
http://www.youtube.com/watch?v=gv8h9Ivs__Y&feature=fvwrel

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If the p-value is greater than alpha then you fail to reject the null hypothesis.

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(-4,4),(-7,3),(-8,2)

Substitute those three points for (x,y) into the standard 
equation for a circle:

(x-h)² + (y-k)² = r²

So we have this system of three equations in three unknowns:

(-4-h)² + (4-k)² = r²
(-7-h)² + (3-k)² = r²
(-8-h)² + (2-k)² = r²

Set the left sides of the first two equations equal:

(-4-h)² + (4-k)² = (-7-h)² + (3-k)² 

Rearrange so as to have difference of squares on each side:

(-4-h)² - (-7-h)² = (3-k)² - (4-k)²

Factor both sides as the difference of squares:

[(-4-h) - (-7-h)][(-4-h) + (-7-h)] = [(3-k) - (4-k)][(3-k) + (4-k)]

Remove the inner parentheses:

[-4 - h + 7 + h][-4 - h - 7 - h] = [3 - k - 4 + k][3 - k + 4 - k]

Simplify:

[3][-11 - 2h] = [-1][7 - 2k]

-33 -6h = -7 + 2k

-6h - 2k = 26

------------------------------------

Set the left sides of the first and third equations equal:

(-4-h)² + (4-k)² = (-8-h)² + (2-k)² 

Rearrange so as to have difference of squares on each side:

(-4-h)² - (-8-h)² = (2-k)² - (4-k)²

Factor both sides as the difference of squares:

[(-4-h) - (-8-h)][(-4-h) + (-8-h)] = [(2-k) - (4-k)][(2-k) + (4-k)]

Remove the inner parentheses:

[-4 - h + 8 + h][-4 - h - 8 - h] = [2 - k - 4 + k][2 - k + 4 - k]

Simplify:

[4][-12 - 2h] = [-2][6 - 2k]

-48 - 8h = -12 + 4k

-8h - 4k = 36

----------------------------------

Now we have the system of equations:

-6h - 2k = 26
-8h - 4k = 36

Solve that system and get (h,k) = (-4,-1)

That's the center.  That's all you asked for.

But if you had been asked for the equation, or if you
wanted to check your problem, you would also need to 
find the radius. 

Substitute (h,k) = (-4,-1) in any one of the original 
3 equations.  I'll pick the first one:

      (-4-h)² + (4-k)² = r²
(-4-(-4))² + (4-(-1))² = r²
      (-4+4)² + (4+1)² = r²
               0² + 5² = r²
                    5² = r²
                     5 = r

So the radius is 5, and so the equation of the circle is:

(x+4)² + (y+1)² = 5²

(x+4)² + (y+1)² = 25

Edwin

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In a class, 25 students have been on a plane, 20 on a train, 27 on a boat, 9 on a plane and train, 12 on a train and boat, 14 on a plane and boat, 3 on all three, and 1 on none of them How many students are in the class?

Let P = the set of all students who have been on a plane, regardless
of whether they have been on a train or a boat.
Let T = the set of all students who have been on a train, regardless
of whether they have been on a plane or a boat.
Let B = the set of all students who have been on a boat, regardless
of whether they have been on a plane or a train.


There are 8 regions which I have labeled with arbitrary small
letters. [I skipped little "b" because I used capital "B" for the
set of all students who have been on a boat.]



The most inclusive clue we have is this one:

>>...3 on all three,...<<

Those 3 are in all three circles, and the only
region that is common to all three circles is
the middle region f, so we put 3 for f



The most EXcluding clue we have is this one:

>>...and 1 on none of them...<<

That 1 is outside all three circles, and the only
region outside all three circles is region i, so we 
put 1 in region i:



Now we look at this:

>>...9 on a plane and train,...<<

3 of those 9 are already accounted for in the middle region,
so that leaves 9-3 or 6 to go in the other region common
to circles P and T, which is region c, so we put 6 in that region:



Now we look at this:

>>...12 on a train and boat,...<<

3 of those 12 are already accounted for in the middle region,
so that leaves 12-3 or 9 to go in the other region common
to circles T and B, which is region g, so we put 9 in that region:



Now we look at this:

>>...14 on a plane and boat,...<<

3 of those 14 are already accounted for in the middle region,
so that leaves 14-3 or 11 to go in the other region common
to circles P and B, which is region e, so we put 11 in that region:


Now we look at this:

>>...25 students have been on a plane,...<<

We have numbers in 3 of the regions of circle P, so 
6+3+11 or 20 of those 25 are already accounted
for so that leaves 25-20 or 5 to go in the remaining
region of circle P, region a, so we put 5 in that region:



Now we look at this:

>>...20 on a train,...<<

We have numbers in 3 of the regions of circle T, so 
6+3+9 or 18 of those 20 are already accounted
for so that leaves 20-18 or 2 to go in the remaining
region of circle T, region d, so we put 2 in that region:


Now we look at this:

>>...27 on a boat,...<<

We have numbers in 3 of the regions of circle B, so 
11+3+9 or 23 of those 27 are already accounted
for so that leaves 27-23 or 4 to go in the remaining
region of circle B, region h, so we put 4 in that region:


We are now ready to look at the question, for now
we can answer anything we are asked because we have
the complete Venn diagram:

>>...How many students are in the class?...<<

Just add up all the 8 numbers in the regions:

5+6+2+11+3+9+4+1 = 41

Answer: There are 41 students in the class.

Edwin

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A RATIOnal number is the RATIO of two integers.

2.86 means  which means  which is the RATIO
of two integers 286 and 100.

[And yes, although irrelevant, it reduces to  which is also
the RATIO of two integers.]

A RATIOnal number is the RATIO of two integers,
therefore 2.86 is a RATIOnal number.
Edwin

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zoe the goat is tied to a rope to one corner of a 15 meter by 25 meter rectangular barn. over what area of the field can zoe graze is the rope is
1. 10 meters long?
2. 20 meters long?
3. 30 meters long?
4. zoe is happiest when she has at least 400 meters squared to graze. what possible lengths of rope could be used?

1. 10 meters long?

That's 3/4ths of a circle with a radius of 10.

The area of a full circle with radius 10 is

A =  =  =  

Multiply that by 3/4ths gives · =  = 235.6 m² 

---------------------------------------------

2. 20 meters long?

That's 3/4ths of a circle with a radius of 20. plus 1/4th of a circle
with a radius of 5

The area of a full circle with radius 20 is

A =  =  =  

Multiply that by 3/4ths gives · =  = 942.48 m² 

The area of a full circle with radius 5 is

A =  =  =  

Multiply that by 1/4th gives · =  = 19.63 m²

Total area = 942.48+19.63 = 962.11 m².

-----------------------------------------------------------

3. 30 meters long?

That's 3/4ths of a circle with a radius of 30.

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A probability has a numerator and a denominator.  The denominator
is easy.  6 ways for the 1st roll times 6 ways for the 2nd roll 
times 6 ways the 3rd roll.  That's 6·6·6 = 6³ = 216

The numerator is not so easy.  The smallest sum the 2nd and 3rd rolls
can have is 2, and the largest sum the 2nd and 3rd rolls can have is 5,
when the 1st roll is 6.  So we make a chart:
    A               B           C             D            E           F 
               Possible       Number      Possible    Number of     Product 
 Sum of        2nd & 3rd      of ways     1st rolls    possible       of 
2nd & 3rd       rolls to      to get      for that     1st rolls    columns  
  rolls       get that sum    that sum     sum       for that sum    C & E 
---------------------------------------------------------------------------
    2              1+1          1         3,4,5,6          4           4
    3            1+2,2+1        2          4,5,6           3           6
    4          1+3,2+2,3+1      3           5,6            2           6
    5        1+4,2+3,3+2,4+1    4            6             1           4
---------------------------------------------------------------------------
                                                              Total = 20

So the numerator of the probability is 20.

Therefore the desired probability is 20 out of 216 or  which
reduces to 

Edwin

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(x² + x - 7)(x + 2)

x²(x + 2) + x(x + 2) - 7(x + 2)

x³ + 2x² + x² + 2x - 7x - 14

x³ + 3x² - 5x - 14

To check, pick an arbitrary value for x (other than 0 or 1).

I pick x=3

Substitute x=3 in the original expression:

(x² + x - 7)(x + 2)
(3² + 3 - 7)(3 + 2)
(9 + 3 - 7)(5)
(12-7)(5)
(5)(5)
25

Now substitute x=3 it in the final answer:

x³ + 3x² - 5x - 14
3³ + 3(3)² - 5(3) - 14
27 + 3(9) - 15 - 14
27 + 27 - 15 - 14
54 - 15 - 14
39 - 14
25

Since both answers are the same, we are pretty sure we are right.
In some rare cases we could still have made a mistake, but it would be
very rare.  To be even surer, pick another number and substitute.

Edwin