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solver91311 answered: 16882 problems
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They have different sized posters of animals in the park. All the posters in the shop are proportional. The smallest poster is 12 inches wide & 15 inches tall. The largest poster is 32 inches wide. How tall is the largest poster? 1 solutions
Answer 412525 by solver91311(16897) on 2012-10-07 16:20:44 (Show Source):
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absolute-value/662523: My math teacher gave us homework for the weekend and I understand the majority of it, but there are a few problems that I just don't understand. I was wondering if I left the problem, someone could show how to solve it.
Math problem:
|m|
---=3
5 1 solutions
Answer 412330 by solver91311(16897) on 2012-10-06 15:54:30 (Show Source):
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Linear-equations/662513: -x+3y=4
x-3y=3
what is the elimination method to solve these two problems 1 solutions
Answer 412323 by solver91311(16897) on 2012-10-06 15:20:30 (Show Source):
You can put this solution on YOUR website!
It is not two problems. It is two equations each of which is in two variables that together form one 2X2 system of linear equations.
Elimination Method.
1. Multiply one or both equations by a constant or constants such that the coefficient on one of the variables is the additive inverse of the coefficient on the same variable in the other equation. (For this system, you can skip this step since the coefficients on are already 1 and -1 and the coefficients on are already 3 and -3)
2. Add the two equations term by term. One (at least) of the variables will be eliminated leaving you either with a single equation in a single variable, a triviality (such as 0 = 0), or an absurdity (such as 0 = 4).
3. If a single equation in a single variable remains, solve it. Use the value you discover for that variable to substitute back into either of the original equations. That gives you a new single equation in the eliminated variable which you must now solve. The two values for the two variables form an ordered pair that is the solution set of the 2X2 system. Graphically, this is represented by two distinct lines that intersect in a single point.
4. If you end up with a triviality, then the solution set of the system is the solution set of either (and therefore both) of the two equations. Graphically, this is represented by each of the equations graphing to the same line.
5. If you end up with an absurdity, then the solution set of the system is the empty set. Graphically, this is represented by two parallel lines.
John

My calculator said it, I believe it, that settles it
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Money_Word_Problems/662502: You find 8 coins in an old desk drawer, consisting entirely of nickels, dimes, and quarters, with a face value of 80 cents. However, the coins all date from 1889 and are worth much more than their face value. You were able to sell all the coins to a coin shop for a total of $22. Suppose you received $3 for each nickel, $2 for each dime, and $5 for each quarter. How many of the coins were quarters? 1 solutions
Answer 412313 by solver91311(16897) on 2012-10-06 14:41:00 (Show Source):
You can put this solution on YOUR website!
Let represent the number of nickels, represent the number of dimes, and represent the number of quarters.
because the total number of coins is 8
because the total FACE value of the coins is 80 cents, each nickel is worth 5 cents, each dime is worth 10 cents, and each quarter is worth 25 cents.
because the total SALE value of the coins is 22 dollars, each nickel is worth 3 dollars, each dime is worth 2 dollars, and each quarter is worth 5 dollars.
Solve the 3X3 system. Report the value of
John

My calculator said it, I believe it, that settles it
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Polynomials-and-rational-expressions/662478: Iam trying to find all the zeros for
x5+6x4-x3-6x2-20x-120 found -6 1 solutions
Answer 412286 by solver91311(16897) on 2012-10-06 12:48:39 (Show Source):
You can put this solution on YOUR website!
Having found -6 is a good start, and if you continued looking for rational roots from there, you were doomed to failure; there aren't any. The other four zeros consist of a pair of irrationals and a conjugate pair of complex numbers. The thing is, after having found the factor , you were staring at a 4th degree polynomial that we now know has no rational factors. Woe is me! What to do now?
Fortunately, this particular 4th degree polynomial is so configured that you can actually treat it like a quadratic. Read on.
First let's do the synthetic division with -6:
-6 | 1 6 -1 -6 -20 -120
-6 0 6 0 120
1 0 -1 0 20 0
From this we can determine that and are factors of , and that -6 is indeed a real rational zero of the 5th degree polynomial.
But what do we do with the 4th degree factor? We use a substitution trick. Let and substitute into :
Et voilą! We have a quadratic that factors tidily:
So the zeros are
and
But wait! We aren't done. The problem doesn't want values of that make the original polynomial equal to zero, it wants values of . So reverse the substitution, thus:
hence
OR
hence
And we are done. We started with a 5th degree polynomial and found 5 zeros satisfying the Fundamental Theorem of Algebra. Time to sit around in a circle, hold hands, and sing "Kumbaya."
John

My calculator said it, I believe it, that settles it
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Equations/662327: When solving a system of equations using the substitution menthod, you obtain the following result: 8 = 9. This result gives you some information about the solution.
a) How many solutions does the system have?
b) What would the graph of the system look like? 1 solutions
Answer 412103 by solver91311(16897) on 2012-10-05 17:50:56 (Show Source):
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Polynomials-and-rational-expressions/662316: Factor completely, if the polynomial is prime, state this. 25a^a-23a-2. I do not understand these what so ever! I have tried using youtube to find videos and they always used little numbers or even. Can someone PLEASE HELP ME! You don't even have to tell me the answer just how to get the answer. 1 solutions
Answer 412102 by solver91311(16897) on 2012-10-05 17:49:39 (Show Source):
You can put this solution on YOUR website!
You really mean , then put a fork in yourself -- you are done. You aren't factoring this anytime in your lifetime.
On the other hand if you were just being sloppy and meant , then read on.
You need to find two factors (you know it is only two factors because the exponent is 2) of the form and so that , , and
Quadratic trinomials come with three signs to consider, that is the sign on each of the three terms. In your case the sign on the first term is positive, and the sign on the other two is negative.
In order to get a negative sign on the constant term, the signs in the middle of each of your binomial factors must be OPPOSITE -- hence either is a negative number or is a negative number.
Now, let's get down to numbers. There are exactly three possibilities for the product if we are dealing with only integer coefficients, namely 1 and 25, 5 and 5, or 25 and 1. There are only 2 possibilities to get the 2 as a constant term, 1 and 2 or 2 and 1.
Next, let's look at the linear or first degree term (the term in the middle of yor trinomial). We need -23, right. Well -25 plus 2 is exactly -23, isn't it?
So, how to make -25? Either -25 times 1 or 25 times -1. So let's try and .
Then if we let and , we end up with -25a plus 2a for the center term.
Use those numbers and FOIL it to check the work.
There are methods for systematically working these out, but I think they are a bit beyond where I think you are in your mathematics education. If you are interested, look up the Rational Roots Theorem and learn how to perform Synthetic Division -- you'll need both.
If you have a graphing calculator, then set your quadratic trinomial equal to zero and submit the equation to your calculator. If the calculator provides rational number zeros, then the quadratic is factorable, otherwise not. Furthermore, the factors are and where and are the zeros of the function. If you end up with the zero being a rational number, i.e. , then instead of writing your factor as , you should write it .
John

My calculator said it, I believe it, that settles it
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