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 Probability-and-statistics/149898: This is an intro to probability chapter in a book called Applied mathmatics. The book is a combo of two diferent books especially made for my school. This is the problem: Find (2n+2)!/(2n)! I know how to do algebra, my problem I believe is because of the ! symbol. I have just been introduced to it and understand the concept , but not how it works in an agebraic expression. Itried soving for n but don't know what to do with the ! ineed to understand the steps involved. this is the answer: 4n2 +6n + 2 thanks in advance for any help. heidi 1 solutions Answer 110003 by jim_thompson5910(28593)   on 2008-07-27 13:24:04 (Show Source): You can put this solution on YOUR website!The notation tells you to simply multiply that number by every integer that is less than that number until you hit 1. So if , this means that . In this case, which simplifies to and So Highlight the common terms. Cancel out the common terms. Simplify. FOIL to get
 Quadratic_Equations/149843: This question is from textbook I don't know how to do this at all be if you can show me then it will help me do the other promblems. 5t^-10t=01 solutions Answer 109970 by jim_thompson5910(28593)   on 2008-07-26 22:13:14 (Show Source): You can put this solution on YOUR website! Start with the given equation Factor the left side Now set each factor equal to zero: or or Now solve for t in each case So our answers are or
 Polynomials-and-rational-expressions/149836: How do you factor xy - 3x - 8y + 24, whats the answer?1 solutions Answer 109964 by jim_thompson5910(28593)   on 2008-07-26 21:00:18 (Show Source): You can put this solution on YOUR website! Start with the given expression Group like terms Factor out the GCF out of the first group. Factor out the GCF out of the second group Since we have the common term , we can combine like terms So factors to
 Polynomials-and-rational-expressions/149833: How do you factor ?1 solutions Answer 109961 by jim_thompson5910(28593)   on 2008-07-26 20:48:39 (Show Source): You can put this solution on YOUR website!Well you can't solve it (since it's not an equation), but you can factor Start with the given expression. Rearrange the terms Group like terms Factor out the GCF out of the first group. Factor out the GCF out of the second group Since we have the common term , we can combine like terms So factors to
 Graphs/149830: 2.) Compound interest. Suppose that $750 is invested at 7% interest, compounded semiannually. A.) Find the function for the amount to which the investment grows after t years. B.)Find the amount of money in the account at t=1, 6,10,15, and 25 yr 1 solutions Answer 109956 by jim_thompson5910(28593) on 2008-07-26 19:52:58 (Show Source): You can put this solution on YOUR website!A) The compound interest formula is where A is the return, P is the principal, r is the interest rate, n is the compound frequency, and t is the time in years. So in this case, the principal is , the rate is (note 7% is 0.07 in decimal form) and the compound frequency is (note: semiannually means that it is compounded twice a year) So the equation is ============================================================================== B) Let's find out how much money there is in the account after 1 year Start with given equation Plug in t=1 Divide 0.07 by 2 to get 0.035 Multiply the exponents 2 and 1 to get 2 Add 1 and 0.035 to get 1.035 Raise 1.035 to the 2 th power to get 1.071225 Multiply 750 and 1.071225 to get 803.41875 So in 1 year, there is$803.42 (rounded to the nearest cent) in the account ------------------------------------------------------------ Let's find out how much money there is in the account after 6 years Start with given equation. Plug in t=6 Divide 0.07 by 2 to get 0.035 Multiply the exponents 2 and 6 to get 12 Add 1 and 0.035 to get 1.035 Raise 1.035 to the 12 th power to get 1.51106865734636 Multiply 750 and 1.51106865734636 to get 1133.30149300977 So in 6 years, there is $1133.30 (rounded to the nearest cent) in the account ------------------------------------------------------------ Let's find out how much money there is in the account after 10 years Start with given equation. Plug in t=10 Divide 0.07 by 2 to get 0.035 Multiply the exponents 2 and 10 to get 20 Add 1 and 0.035 to get 1.035 Raise 1.035 to the 20 th power to get 1.98978886346584 Multiply 750 and 1.98978886346584 to get 1492.34164759938 So in 10 years, there is$1492.34 (rounded to the nearest cent) in the account ------------------------------------------------------------ Let's find out how much money there is in the account after 15 years Start with given equation. Plug in t=15 Divide 0.07 by 2 to get 0.035 Multiply the exponents 2 and 15 to get 30 Add 1 and 0.035 to get 1.035 Raise 1.035 to the 30 th power to get 2.80679370470263 Multiply 750 and 2.80679370470263 to get 2105.09527852697 So in 15 years, there is $2105.10 (rounded to the nearest cent) in the account ------------------------------------------------------------ Let's find out how much money there is in the account after 15 years Start with given equation. Plug in t=25 Divide 0.07 by 2 to get 0.035 Multiply the exponents 2 and 25 to get 50 Add 1 and 0.035 to get 1.035 Raise 1.035 to the 50 th power to get 5.58492685566332 Multiply 750 and 5.58492685566332 to get 4188.69514174749 So in 25 years, there is$4188.69 (rounded to the nearest cent) in the account
 Graphs/149829: Solve the inequality x^2-x-6>01 solutions Answer 109955 by jim_thompson5910(28593)   on 2008-07-26 19:30:43 (Show Source): You can put this solution on YOUR website! Start with the given inequality Factor the left side Set the left side equal to zero Set each individual factor equal to zero: or Solve for x in each case: or So our critical values are and Now set up a number line and plot the critical values on the number line So let's pick some test points that are near the critical values and evaluate them. Let's pick a test value that is less than (notice how it's to the left of the leftmost endpoint): So let's pick Start with the given inequality Plug in Evaluate and simplify the left side Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set. So part our solution in interval notation is () --------------------------------------------------------------------------------------------- Let's pick a test value that is in between and : So let's pick Start with the given inequality Plug in Evaluate and simplify the left side Since the inequality is false, this means that the interval does not work. So this interval is not in our solution set and we can ignore it. --------------------------------------------------------------------------------------------- Let's pick a test value that is greater than (notice how it's to the right of the rightmost endpoint): So let's pick Start with the given inequality Plug in Evaluate and simplify the left side Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set. So part our solution in interval notation is () --------------------------------------------------------------------------------------------- Summary: So the solution in interval notation is: () () Here's a graph to visually verify the answer
 Exponents/149819: Find the exact area of a triangle with a base of sqrt of 30 meters and height sqrt of 6 meters.1 solutions Answer 109932 by jim_thompson5910(28593)   on 2008-07-26 12:35:57 (Show Source): You can put this solution on YOUR website! Start with the area of a triangle formula Plug in and Multiply Simplify the square root. Reduce to get or just 3 So the area is
 Radicals/149817: Is sqrt of a + sqrt of b = sqrt a+b for all values of a and b?1 solutions Answer 109929 by jim_thompson5910(28593)   on 2008-07-26 12:33:03 (Show Source): You can put this solution on YOUR website!Let's assume that for all values of a and b. So now let's just pick arbitrary values for a and b. So let's make and . Start with the given equation. Plug in and Add 2 and 3 to get 5 Take the square root of 2 to get 1.41421. Take the square root of 3 to get 1.73205. Take the square root of 5 to get 2.23607. Add 1.41421 and 1.73205 to get 3.14626 Since , this shows us that . So this means that . There are some values that will make it true. For instance and are one pair of values, and are another. However, in general, the equation is not true.
 Exponents/149810: A 30 inch by 40 inch countertop for a work island is to be covered with green ceramic tiles, except for a border of uniform width, if the area covered by green tiles is 704 square inches (in^2 ) , then how wide is the border?1 solutions Answer 109925 by jim_thompson5910(28593)   on 2008-07-26 12:21:38 (Show Source): You can put this solution on YOUR website!Let x=width of border. First, let's draw the picture. note: the lengths and are due to the subtraction of 2 "x" lengths. Take note that there are two dashed lines per side that are not part of the green tiles. From the picture, we see that the actual counter top is by . So the length is and the width is Now remember, the area of a rectangle is Plug in the given area of the green tiles , and FOIL Subtract 704 from both sides. Combine like terms. Let's use the quadratic formula to solve for x Start with the quadratic formula Plug in , , and Negate to get . Square to get . Multiply to get Subtract from to get Multiply and to get . Take the square root of to get . or Break up the expression. or Combine like terms. or Simplify. So the possible answers are or However, we need to see if they generate reasonable lengths and widths Let's check the solution Go back to the length equation Plug in Multiply Subtract Since a negative length is not possible, this means that the value is a reasonable solution. ---------------------------- Let's check the solution Go back to the length equation Plug in Multiply Subtract So we get a reasonable length with -------- Go back to the width equation Plug in Multiply Subtract So we also get a reasonable width with =============================================== Answer: So the only solution is So the width of the border is 4 inches
Polynomials-and-rational-expressions/149802: 1. Use synthetic division to divide the polynomial 2x^3 + 7x^2 – 5 by 3+x and write the quotient and the remainder
2. Consider the polynomial f(x)= 4x^4+5x^3 + 7x^2 – 34x+8
By using the Rational Zero Theorem, list all possible rational zeros of the given polynomial.

Find all of the zeros of the given polynomial. Show procedure

1 solutions

Answer 109921 by jim_thompson5910(28593)   on 2008-07-26 11:37:26 (Show Source):
You can put this solution on YOUR website!
# 1

Let's simplify this expression using synthetic division

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is -3

Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from to there is a zero coefficient for . This is simply because really looks like
 -3 | 2 7 0 -5 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
 -3 | 2 7 0 -5 | 2

Multiply -3 by 2 and place the product (which is -6) right underneath the second coefficient (which is 7)
 -3 | 2 7 0 -5 | -6 2

Add -6 and 7 to get 1. Place the sum right underneath -6.
 -3 | 2 7 0 -5 | -6 2 1

Multiply -3 by 1 and place the product (which is -3) right underneath the third coefficient (which is 0)
 -3 | 2 7 0 -5 | -6 -3 2 1

Add -3 and 0 to get -3. Place the sum right underneath -3.
 -3 | 2 7 0 -5 | -6 -3 2 1 -3

Multiply -3 by -3 and place the product (which is 9) right underneath the fourth coefficient (which is -5)
 -3 | 2 7 0 -5 | -6 -3 9 2 1 -3

Add 9 and -5 to get 4. Place the sum right underneath 9.
 -3 | 2 7 0 -5 | -6 -3 9 2 1 -3 4

Since the last column adds to 4, we have a remainder of 4. This means is not a factor of

Now lets look at the bottom row of coefficients:

The first 3 coefficients (2,1,-3) form the quotient

and the last coefficient 4, is the remainder, which is placed over like this

Putting this altogether, we get:

So

which looks like this in remainder form:
remainder 4

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

So the quotient is and the remainder is 4

# 2

a)

Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients

So let's list the factors of 8 (the last coefficient):

Now let's list the factors of 4 (the first coefficient):

Now let's divide each factor of the last coefficient by each factor of the first coefficient

Now simplify

These are all the distinct rational zeros of the function that could occur

b)

Now let's use synthetic division to test each possible zero:

Let's make the synthetic division table for the function given the possible zero :
 1/2 | 4 5 7 -34 8 | 2 7/2 21/4 -115/8 4 7 21/2 -115/4 -51/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 1/4 | 4 5 7 -34 8 | 1 3/2 17/8 -255/32 4 6 17/2 -255/8 1/32

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 2 | 4 5 7 -34 8 | 8 26 66 64 4 13 33 32 72

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 4 | 4 5 7 -34 8 | 16 84 364 1320 4 21 91 330 1328

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 8 | 4 5 7 -34 8 | 32 296 2424 19120 4 37 303 2390 19128

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -1 | 4 5 7 -34 8 | -4 -1 -6 40 4 1 6 -40 48

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -1/2 | 4 5 7 -34 8 | -2 -3/2 -11/4 147/8 4 3 11/2 -147/4 211/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -1/4 | 4 5 7 -34 8 | -1 -1 -3/2 71/8 4 4 6 -71/2 135/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -2 | 4 5 7 -34 8 | -8 6 -26 120 4 -3 13 -60 128

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -4 | 4 5 7 -34 8 | -16 44 -204 952 4 -11 51 -238 960

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -8 | 4 5 7 -34 8 | -32 216 -1784 14544 4 -27 223 -1818 14552

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

====================================

Since none of the possible rational roots are actual roots, this means that the polynomial either has irrational roots or complex roots.

 Trigonometry-basics/149789: find the exact value of tan(-3 pie)1 solutions Answer 109891 by jim_thompson5910(28593)   on 2008-07-25 21:26:47 (Show Source): You can put this solution on YOUR website! Use the identity to rewrite the expression. Rewrite tangent in terms of sine and cosine. Rewrite as Rewrite the expression using the identities and Now, let's reference the unit circle From the unit circle, we can see that at the angle , there is a point on the unit circle . So this tells us that and . Also at the angle , there is a point on the unit circle . So this tells us that and . Take the cosine of to get -1. Take the sine of to get 0. Take the cosine of to get 1. Take the sine of to get 0. Multiply Add Reduce So
Polynomials-and-rational-expressions/149769: don't know how to check my amswers for these....
simplify: (3x^3-x^2-4x+1)/(x-1)
factor: 18x+x^2-11x
factor: 6x^2+x-2
factor: 12x^3+31x^2+20x
1 solutions

Answer 109869 by jim_thompson5910(28593)   on 2008-07-25 17:08:18 (Show Source):
You can put this solution on YOUR website!
# 1

Let's simplify this expression using synthetic division

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is 1

Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
 1 | 3 -1 -4 1 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 3)
 1 | 3 -1 -4 1 | 3

Multiply 1 by 3 and place the product (which is 3) right underneath the second coefficient (which is -1)
 1 | 3 -1 -4 1 | 3 3

Add 3 and -1 to get 2. Place the sum right underneath 3.
 1 | 3 -1 -4 1 | 3 3 2

Multiply 1 by 2 and place the product (which is 2) right underneath the third coefficient (which is -4)
 1 | 3 -1 -4 1 | 3 2 3 2

Add 2 and -4 to get -2. Place the sum right underneath 2.
 1 | 3 -1 -4 1 | 3 2 3 2 -2

Multiply 1 by -2 and place the product (which is -2) right underneath the fourth coefficient (which is 1)
 1 | 3 -1 -4 1 | 3 2 -2 3 2 -2

Add -2 and 1 to get -1. Place the sum right underneath -2.
 1 | 3 -1 -4 1 | 3 2 -2 3 2 -2 -1

Since the last column adds to -1, we have a remainder of -1. This means is not a factor of
Now lets look at the bottom row of coefficients:

The first 3 coefficients (3,2,-2) form the quotient

and the last coefficient -1, is the remainder, which is placed over like this

Putting this altogether, we get:

So

Combine like terms.

Factor out the GCF

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

So factors to

# 3

Looking at we can see that the first term is and the last term is where the coefficients are 6 and -2 respectively.

Now multiply the first coefficient 6 and the last coefficient -2 to get -12. Now what two numbers multiply to -12 and add to the middle coefficient 1? Let's list all of the factors of -12:

Factors of -12:
1,2,3,4,6,12

-1,-2,-3,-4,-6,-12 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to -12
(1)*(-12)
(2)*(-6)
(3)*(-4)
(-1)*(12)
(-2)*(6)
(-3)*(4)

note: remember, the product of a negative and a positive number is a negative number

Now which of these pairs add to 1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 1

First NumberSecond NumberSum
1-121+(-12)=-11
2-62+(-6)=-4
3-43+(-4)=-1
-112-1+12=11
-26-2+6=4
-34-3+4=1

From this list we can see that -3 and 4 add up to 1 and multiply to -12

Now looking at the expression , replace with (notice adds up to . So it is equivalent to )

Now let's factor by grouping:

Group like terms

Factor out the GCF of out of the first group. Factor out the GCF of out of the second group

Since we have a common term of , we can combine like terms

So factors to

So this also means that factors to (since is equivalent to )

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

So factors to

# 4

Factor out the GCF

Now let's focus on the inner expression

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

Looking at we can see that the first term is and the last term is where the coefficients are 12 and 20 respectively.

Now multiply the first coefficient 12 and the last coefficient 20 to get 240. Now what two numbers multiply to 240 and add to the middle coefficient 31? Let's list all of the factors of 240:

Factors of 240:
1,2,3,4,5,6,8,10,12,15,16,20,24,30,40,48,60,80,120,240

-1,-2,-3,-4,-5,-6,-8,-10,-12,-15,-16,-20,-24,-30,-40,-48,-60,-80,-120,-240 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to 240
1*240
2*120
3*80
4*60
5*48
6*40
8*30
10*24
12*20
15*16
(-1)*(-240)
(-2)*(-120)
(-3)*(-80)
(-4)*(-60)
(-5)*(-48)
(-6)*(-40)
(-8)*(-30)
(-10)*(-24)
(-12)*(-20)
(-15)*(-16)

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 31? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 31

First NumberSecond NumberSum
12401+240=241
21202+120=122
3803+80=83
4604+60=64
5485+48=53
6406+40=46
8308+30=38
102410+24=34
122012+20=32
151615+16=31
-1-240-1+(-240)=-241
-2-120-2+(-120)=-122
-3-80-3+(-80)=-83
-4-60-4+(-60)=-64
-5-48-5+(-48)=-53
-6-40-6+(-40)=-46
-8-30-8+(-30)=-38
-10-24-10+(-24)=-34
-12-20-12+(-20)=-32
-15-16-15+(-16)=-31

From this list we can see that 15 and 16 add up to 31 and multiply to 240

Now looking at the expression , replace with (notice adds up to . So it is equivalent to )

Now let's factor by grouping:

Group like terms

Factor out the GCF of out of the first group. Factor out the GCF of out of the second group

Since we have a common term of , we can combine like terms

So factors to

So this also means that factors to (since is equivalent to )

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

So our expression goes from and factors further to

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

So factors to

Rational-functions/149768: Graph the polynomial function to approximately find the function's zeros, then use synthetic division and the remainder theorem to exactly find its zeros.
1 solutions

Answer 109867 by jim_thompson5910(28593)   on 2008-07-25 16:37:14 (Show Source):
You can put this solution on YOUR website!
First, let's graph the function to get

From the graph, we can see that the graph has the approximate zeros and

==============================================================

Now let's use the Rational Root Theorem to list all of the possible rational roots

Rational Root Theorem:

where p and q are the factors of the last and first coefficients

So let's list the factors of -2 (the last coefficient):

Now let's list the factors of 1 (the first coefficient):

Now let's divide each factor of the last coefficient by each factor of the first coefficient

Now simplify

These are all the distinct rational zeros of the function that could occur

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

Now let's use synthetic division to test each possible zero

Let's see if the possible zero is really a root for the function

So let's make the synthetic division table for the function given the possible zero :
 1 | 1 1 -3 -5 -2 | 1 2 -1 -6 1 2 -1 -6 -8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's see if the possible zero is really a root for the function

So let's make the synthetic division table for the function given the possible zero :
 2 | 1 1 -3 -5 -2 | 2 6 6 2 1 3 3 1 0

Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of

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

Let's see if the possible zero is really a root for the function

So let's make the synthetic division table for the function given the possible zero :
 -1 | 1 1 -3 -5 -2 | -1 0 3 2 1 0 -3 -2 0

Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of

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

Let's see if the possible zero is really a root for the function

So let's make the synthetic division table for the function given the possible zero :
 -2 | 1 1 -3 -5 -2 | -2 2 2 6 1 -1 -1 -3 4

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

===========================================
Summary:

So only and are actually rational roots.

Now looking back at the table for the test zero , we see
 -1 | 1 1 -3 -5 -2 | -1 0 3 2 1 0 -3 -2 0

The bottom row of coefficients (minus the last one) form the quotient

Now let's perform synthetic division using the other zero on the function

 2 | 1 0 -3 -2 | 2 4 2 1 2 1 0

Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of

Once again the first three coefficients in the bottom row form the quotient

Let's use the quadratic formula to find the zeros of

Plug in , , and

Square to get .

Multiply to get

Subtract from to get

Multiply and to get .

Take the square root of to get .

or Break up the expression.

or Combine like terms.

or Simplify.

So the zeros of are or or just with a multiplicity of 2

Now there are 3 instances where we get a zero of . So this tells us that the zero has a multiplicity of 3

================================================

So the zeros of are (with a multiplicity of 3) and

Rational-functions/149767: What is the remainder when is divided by
1 solutions

Answer 109866 by jim_thompson5910(28593)   on 2008-07-25 16:18:09 (Show Source):
You can put this solution on YOUR website!
Let's simplify this expression using synthetic division

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is 1

Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from to there is a zero coefficient for , , and . This is simply because really looks like
 1 | 1 -1 1 -1 0 0 0 4 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
 1 | 1 -1 1 -1 0 0 0 4 | 1

Multiply 1 by 1 and place the product (which is 1) right underneath the second coefficient (which is -1)
 1 | 1 -1 1 -1 0 0 0 4 | 1 1

Add 1 and -1 to get 0. Place the sum right underneath 1.
 1 | 1 -1 1 -1 0 0 0 4 | 1 1 0

Multiply 1 by 0 and place the product (which is 0) right underneath the third coefficient (which is 1)
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0

Add 0 and 1 to get 1. Place the sum right underneath 0.
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0 1

Multiply 1 by 1 and place the product (which is 1) right underneath the fourth coefficient (which is -1)
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 1 0 1

Add 1 and -1 to get 0. Place the sum right underneath 1.
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 1 0 1 0

Multiply 1 by 0 and place the product (which is 0) right underneath the fifth coefficient (which is 0)
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0 1 0 1 0

Add 0 and 0 to get 0. Place the sum right underneath 0.
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0 1 0 1 0 0

Multiply 1 by 0 and place the product (which is 0) right underneath the sixth coefficient (which is 0)
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0 0 1 0 1 0 0

Add 0 and 0 to get 0. Place the sum right underneath 0.
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0 0 1 0 1 0 0 0

Multiply 1 by 0 and place the product (which is 0) right underneath the seventh coefficient (which is 0)
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0 0 0 1 0 1 0 0 0

Add 0 and 0 to get 0. Place the sum right underneath 0.
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0 0 0 1 0 1 0 0 0 0

Multiply 1 by 0 and place the product (which is 0) right underneath the eighth coefficient (which is 4)
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0 0 0 0 1 0 1 0 0 0 0

Add 0 and 4 to get 4. Place the sum right underneath 0.
 1 | 1 -1 1 -1 0 0 0 4 | 1 0 1 0 0 0 0 1 0 1 0 0 0 0 4

Since the last column adds to 4, we have a remainder of 4.

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

So when is divided by , the remainder is 4

Rational-functions/149766: Find the quotient of divided by
1 solutions

Answer 109865 by jim_thompson5910(28593)   on 2008-07-25 16:12:43 (Show Source):
You can put this solution on YOUR website!
Let's simplify this expression using synthetic division

First lets find our test zero:

Set the denominator equal to zero

Solve for x.

so our test zero is -2

Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.(note: remember if a polynomial goes from to there is a zero coefficient for , , , and . This is simply because really looks like
 -2 | 1 0 0 0 0 32 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 1)
 -2 | 1 0 0 0 0 32 | 1

Multiply -2 by 1 and place the product (which is -2) right underneath the second coefficient (which is 0)
 -2 | 1 0 0 0 0 32 | -2 1

Add -2 and 0 to get -2. Place the sum right underneath -2.
 -2 | 1 0 0 0 0 32 | -2 1 -2

Multiply -2 by -2 and place the product (which is 4) right underneath the third coefficient (which is 0)
 -2 | 1 0 0 0 0 32 | -2 4 1 -2

Add 4 and 0 to get 4. Place the sum right underneath 4.
 -2 | 1 0 0 0 0 32 | -2 4 1 -2 4

Multiply -2 by 4 and place the product (which is -8) right underneath the fourth coefficient (which is 0)
 -2 | 1 0 0 0 0 32 | -2 4 -8 1 -2 4

Add -8 and 0 to get -8. Place the sum right underneath -8.
 -2 | 1 0 0 0 0 32 | -2 4 -8 1 -2 4 -8

Multiply -2 by -8 and place the product (which is 16) right underneath the fifth coefficient (which is 0)
 -2 | 1 0 0 0 0 32 | -2 4 -8 16 1 -2 4 -8

Add 16 and 0 to get 16. Place the sum right underneath 16.
 -2 | 1 0 0 0 0 32 | -2 4 -8 16 1 -2 4 -8 16

Multiply -2 by 16 and place the product (which is -32) right underneath the sixth coefficient (which is 32)
 -2 | 1 0 0 0 0 32 | -2 4 -8 16 -32 1 -2 4 -8 16

Add -32 and 32 to get 0. Place the sum right underneath -32.
 -2 | 1 0 0 0 0 32 | -2 4 -8 16 -32 1 -2 4 -8 16 0

Since the last column adds to zero, we have a remainder of zero. This means is a factor of

Now lets look at the bottom row of coefficients:

The first 5 coefficients (1,-2,4,-8,16) form the quotient

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

So which means that the quotient is and the remainder is 0.

 Rational-functions/149765: Use the Intermediate Value Theorem to determine if has a zero in the interval [1,2]1 solutions Answer 109864 by jim_thompson5910(28593)   on 2008-07-25 16:09:18 (Show Source): You can put this solution on YOUR website!To use the intermediate value theorem, simply evaluate the endpoints of the interval. Let's evaluate the left endpoint Start with the given equation. Plug in . Raise to the 5th power to get . Multiply and to get . Multiply and to get . Combine like terms. So the function value at is . This makes the point (1,-4) which tells us that the y-coordinate is negative. -------------------------------------- Now let's evaluate the right endpoint Start with the given equation. Plug in . Raise to the 5th power to get . Multiply and to get . Multiply and to get . Combine like terms. So the function value at is . This makes the point (2,51) which tells us that the y-coordinate is positive. Since the sign of the y-coordinate transitioned from negative to positive on the interval [1,2], this means that the graph must have crossed the x-axis somewhere in the interval. So there is a zero in the interval [1,2].
Rational-functions/149763: Find the zeros of and the multiplicity of each.
1 solutions

Answer 109863 by jim_thompson5910(28593)   on 2008-07-25 16:05:48 (Show Source):
You can put this solution on YOUR website!
First, let's list all of the possible rational zeros.

Any rational zero can be found through this formula

where p and q are the factors of the last and first coefficients

So let's list the factors of 2 (the last coefficient):

Now let's list the factors of 1 (the first coefficient):

Now let's divide each factor of the last coefficient by each factor of the first coefficient

Now simplify

These are all the distinct rational zeros of the function that could occur

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

Now let's test each possible rational root with use of synthetic division

Let's see if the possible zero is really a root for the function

So let's make the synthetic division table for the function given the possible zero :
 1 | 1 -1 -2 2 | 1 0 -2 1 0 -2 0

Since the remainder (the right most entry in the last row) is equal to zero, this means that is a zero of

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

Let's see if the possible zero is really a root for the function

So let's make the synthetic division table for the function given the possible zero :
 2 | 1 -1 -2 2 | 2 2 0 1 1 0 2

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's see if the possible zero is really a root for the function

So let's make the synthetic division table for the function given the possible zero :
 -1 | 1 -1 -2 2 | -1 2 0 1 -2 0 2

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's see if the possible zero is really a root for the function

So let's make the synthetic division table for the function given the possible zero :
 -2 | 1 -1 -2 2 | -2 6 -8 1 -3 4 -6

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

===========================================

So to recap, we only found one rational zero. So the only rational zero for the function is

Now if we go back to the corresponding synthetic division table for the test zero , we get

 1 | 1 -1 -2 2 | 1 0 -2 1 0 -2 0

Now looking at the bottom row of coefficients, we see the first three numbers: 1, 0 and -2

These coefficients form the quotient or just

This means that or

Set the quotient equal to zero

Take the square root of both sides.

or Break up the expression.

So the remaining two zeros are or

===============================================

So the zeros of are:

, or where each zero has a multiplicity of 1.

 Miscellaneous_Word_Problems/149757: A wooden fence surrounds a garden that is in the shape of an irregular triangle. If one side is twice as long as the first, and another is half as long as the first side, and the fence is 70 feet long, what are the dimensions of the garden? ( Make a drawing, assign a variable, answer the question asked, check your work) Thank you very much.1 solutions Answer 109862 by jim_thompson5910(28593)   on 2008-07-25 15:33:06 (Show Source): You can put this solution on YOUR website!Let x=length of first side. Since "one side is twice as long as the first", this means that Also, because "another is half as long as the first side", this means that However, there's a problem. In order for any triangle to be constructed, the lengths of any two sides must be greater than the length of the third side. So for instance, this means that . Plug in and Add Since this inequality is never true for any positive x values, this means that you cannot construct a triangle with sides of x, 2x, and (go ahead and try it out on paper if you don't believe me). So I would double check the problem.
 Trigonometry-basics/149721: Find the exact value of cot(-pie/6)1 solutions Answer 109860 by jim_thompson5910(28593)   on 2008-07-25 15:21:18 (Show Source): You can put this solution on YOUR website! Start with the given expression. Rewrite the cotangent in terms of tangent. Use the identity to rewrite the expression. Reduce. Rewrite the tangent function in terms of sine and cosine. Multiply -1 by the reciprocal of the dividing fraction. Multiply. Now, let's reference the unit circle From the unit circle, we can see that at the angle , there is a point on the unit circle . So this tells us that and So this means that then becomes Multiply the first fraction by the reciprocal of the second fraction. Cancel like terms. Multiply and simplify. So
 Polynomials-and-rational-expressions/149748: This question is from textbook Elementary and Intermediate Algebra -2x – 8 x²+ 2x-8 1 solutions Answer 109858 by jim_thompson5910(28593)   on 2008-07-25 15:08:45 (Show Source): You can put this solution on YOUR website! Start with the given expression. Factor to get . Factor to get . Highlight the common terms. Cancel out the common terms. Simplify. So simplifies to . In other words, where or
 Polynomials-and-rational-expressions/149749: This question is from textbook Elementary and Intermediate Algebra Perform the indicated operation -2a² 20a 3a² ˚ 15a³ 1 solutions Answer 109857 by jim_thompson5910(28593)   on 2008-07-25 15:05:05 (Show Source): You can put this solution on YOUR website! Start with the given expression. Combine the fractions. Multiply. Expand. Remember, and Highlight the common terms. Cancel out the common terms. Simplify. Regroup. So simplifies to . In other words, where
 Square-cubic-other-roots/149754: 3rd time lucky maybe I need some help on one of my questions. I need to make g the subject... T = 2(pi)*[square root of (l/g^3)} Can someone please help me......: and its the letter L not the number 11 solutions Answer 109856 by jim_thompson5910(28593)   on 2008-07-25 14:59:24 (Show Source): You can put this solution on YOUR website! Start with the given equation. Divide both sides by . Square both sides. This will eliminate the square root. Multiply both sides by . Distribute the outer exponent. Square to get . Multiply both sides by . Divide both sides by . Take the cube root of both sides.
 logarithm/149752: log r - log t + 2 log s 1 solutions Answer 109855 by jim_thompson5910(28593)   on 2008-07-25 14:53:08 (Show Source): You can put this solution on YOUR website! Start with the given expression. Rewrite as using the identity Combine the logs using the identity Combine the logs using the identity Simplify. So simplifies to
 logarithm/149753: log x + log 2 = 51 solutions Answer 109853 by jim_thompson5910(28593)   on 2008-07-25 14:45:30 (Show Source): You can put this solution on YOUR website! Start with the given equation. Combine the logs using the identity Rewrite the equation using the property: ====> Raise 10 to the 5th power to get 100,000 Divide both sides by 2. So the answer is
Graphs/149741: 1. Given the polynomial f(x) = 4x^4 + 5x^3 + 7x^2 - 34x + 8

(a) By using the Rational Zero Theorem, list all possible rational zeros of the given polynomial.

(b) Find all of the zeros of the given polynomial. Show the procedure

1 solutions

Answer 109852 by jim_thompson5910(28593)   on 2008-07-25 14:32:40 (Show Source):
You can put this solution on YOUR website!
a)

Any rational zero can be found through this equation

where p and q are the factors of the last and first coefficients

So let's list the factors of 8 (the last coefficient):

Now let's list the factors of 4 (the first coefficient):

Now let's divide each factor of the last coefficient by each factor of the first coefficient

Now simplify

These are all the distinct rational zeros of the function that could occur

b)

Now let's use synthetic division to test each possible zero:

Let's make the synthetic division table for the function given the possible zero :
 1/2 | 4 5 7 -34 8 | 2 7/2 21/4 -115/8 4 7 21/2 -115/4 -51/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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Let's make the synthetic division table for the function given the possible zero :
 1/4 | 4 5 7 -34 8 | 1 3/2 17/8 -255/32 4 6 17/2 -255/8 1/32

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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Let's make the synthetic division table for the function given the possible zero :
 2 | 4 5 7 -34 8 | 8 26 66 64 4 13 33 32 72

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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Let's make the synthetic division table for the function given the possible zero :
 4 | 4 5 7 -34 8 | 16 84 364 1320 4 21 91 330 1328

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 8 | 4 5 7 -34 8 | 32 296 2424 19120 4 37 303 2390 19128

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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Let's make the synthetic division table for the function given the possible zero :
 -1 | 4 5 7 -34 8 | -4 -1 -6 40 4 1 6 -40 48

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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Let's make the synthetic division table for the function given the possible zero :
 -1/2 | 4 5 7 -34 8 | -2 -3/2 -11/4 147/8 4 3 11/2 -147/4 211/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -1/4 | 4 5 7 -34 8 | -1 -1 -3/2 71/8 4 4 6 -71/2 135/8

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -2 | 4 5 7 -34 8 | -8 6 -26 120 4 -3 13 -60 128

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -4 | 4 5 7 -34 8 | -16 44 -204 952 4 -11 51 -238 960

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

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

Let's make the synthetic division table for the function given the possible zero :
 -8 | 4 5 7 -34 8 | -32 216 -1784 14544 4 -27 223 -1818 14552

Since the remainder (the right most entry in the last row) is not equal to zero, this means that is not a zero of

====================================

Since none of the possible rational roots are actual roots, this means that the polynomial either has irrational roots or complex roots.

 Inequalities/149690: -2x-4<10 I think i need to get rid of the -4 first. -2x-4<10 -2x+4<10+4 -2x+4<14 not sure where to go from here, nor am i sure this is right1 solutions Answer 109820 by jim_thompson5910(28593)   on 2008-07-25 01:03:14 (Show Source): You can put this solution on YOUR website! Start with the given inequality. Add to both sides. Combine like terms on the right side. Divide both sides by to isolate . note: Remember, the inequality sign flips when we divide both sides by a negative number. Reduce. ---------------------------------------------------------------------- Answer: So the answer is So the answer in interval notation is Also, the answer in set-builder notation is Here's the graph of the solution set
 Expressions-with-variables/149691: 3x + 4y = 8 for y1 solutions Answer 109819 by jim_thompson5910(28593)   on 2008-07-25 01:01:29 (Show Source): You can put this solution on YOUR website! Start with the given equation. Subtract from both sides. Rearrange the terms. Divide both sides by to isolate . Break up the fraction and simplify.
 Inequalities/149694: This question is from textbook Thank you in advance for helping me with writing this problem in interval notation. The absolute value of 2x+1>=3 I have( -infintiy, -2]U[1,infinity ) Am I close? 1 solutions Answer 109818 by jim_thompson5910(28593)   on 2008-07-25 00:59:50 (Show Source): You can put this solution on YOUR website! Start with the given inequality Break up the absolute value (remember, if you have , then or ) or Break up the absolute value inequality using the given rule Now lets focus on the first inequality Start with the given inequality Subtract 1 from both sides Combine like terms on the right side Divide both sides by 2 to isolate x Divide Now lets focus on the second inequality Start with the given inequality Subtract 1 from both sides Combine like terms on the right side Divide both sides by 2 to isolate x Divide ---------------------------------------------------- Answer: So our answer is or So the solution in interval notation is: (] [) Here's the graph of the solution set Note: There is a closed circle at which means that we're including that value from the solution set. Also, there is a closed circle at which means that we're including that value from the solution set.
 Equations/149693: This question is from textbook Thank you in advance for helping me with writing this problem in interval notation. The absolute value of 5x-3=101 solutions Answer 109817 by jim_thompson5910(28593)   on 2008-07-25 00:56:55 (Show Source): You can put this solution on YOUR website! Start with the given equation Break up the absolute value (remember, if you have , then or ) or Set the expression equal to the original value 10 and it's opposite -10 Now lets focus on the first equation Add 3 to both sides Combine like terms on the right side Divide both sides by 5 to isolate x Reduce Now lets focus on the second equation Add 3 to both sides Combine like terms on the right side Divide both sides by 5 to isolate x Reduce So the solutions to are: and
 Inequalities/149696: How do you write an inequality using x Example:i have at least $25 in my pocket 1 solutions Answer 109816 by jim_thompson5910(28593) on 2008-07-25 00:55:05 (Show Source): You can put this solution on YOUR website!Let x=amount of money in your pocket. So when you have "at least$25", this means that you have 25 or more. Mathematically speaking, this means that some unknown quantity "x" is greater than or equal to 25. So symbolically, this inequality is
 Inequalities/149695: This question is from textbook Thank you in advance for helping me with writing this problem in interval notation. The absolute value of 5n-2<2 I have 01 solutions Answer 109815 by jim_thompson5910(28593)   on 2008-07-25 00:52:33 (Show Source): You can put this solution on YOUR website! Start with the given inequality Break up the absolute value (remember, if you have , then and ) and Break up the absolute value inequality using the given rule Combine the two inequalities to get a compound inequality Add 2 to all sides Divide all sides by 5 to isolate n ---------------------------------------------------- Answer: So our answer is which looks like this in interval notation