Guide :

Explain why the product (3x-3)(3x3) is a constant

~~Explain why the product (3x-3)(3x3) is a constant

Research, Knowledge and Information :


Explain why the product (3x^-3)(3x^3) is a constant ...


Explain why the product (3x^-3)(3x^3) is a constant. - 3551093. 1. Log in Join now Katie; ... Explain why the product (3x^-3)(3x^3) is a constant. 1. Ask for details ;
Read More At : brainly.com...

Problems - Biocomputing Unit


1 3x 3 = 8 2x 1 + 2x 2 + 9x 3 = 7 x ... 2 = 3 x 3 = 1 Therefore, ... Explain. Solution: Let us construct the augmented system matrix 0 @ 2 3 1 6 5 0 2 5 7 1 A
Read More At : biocomp.cnb.csic.es...

Math 115 Spring 2014 Written Homework 6-SOLUTIONS Due Friday ...


Math 115 Spring 2014 Written Homework 6-SOLUTIONS ... explain why not. ... x 3 x3 + 3x2 13x 15 3x + 3x2 6x2 13x 6x2 + 18x
Read More At : www.math.uiuc.edu...

Finding the Determinant of a 3 x 3 matrix - YouTube


Mar 11, 2009 · Finding the Determinant of a 3 x 3 matrix. I show the basic formula and compute the determinant of a specific matrix. For more free math videos, visit http ...

Algebra 1/Algebra 2 EOC Study Flashcards | Quizlet


Start studying Algebra 1/Algebra 2 EOC Study. Learn ... find the product of 3x(x^2 + x ... If it is not a polynomial in on variable explain why. 3x^4 + 6x^3 ...
Read More At : quizlet.com...

Linear Equations and Matrices


C H A P T E R 3 Linear Equations and Matrices ... and y is called the constant term of the equation. ... !3x 2 +!!!x 3 =6!!x 1 +5x 2!2x 3 =12
Read More At : cseweb.ucsd.edu...

POLYNOMIALS - California State University San Marcos


... are the constant polynomials a 0. ... the product of 2x2 + 3x + 1 with 3x2 + 2 can be computed ... if f = x3 in Z 12[x] then f(x + 2) = (x + 2)3 = x3 + 6x2 + 8.
Read More At : public.csusm.edu...

í )( - Montville Township School District


The leading coefficient is 6 and the constant term is í60. The possible rational zeros are RU. ... solutions are í1, 2, 3, and í2. x4 ± 3x3 ± 20 x2 + 84 x ± 80 = 0
Read More At : www.montville.net...

Basics of Polynomials - UUMath


Basics of Polynomials ... 0 constant 2 2 2 1 linear 3x1 3 3x 2 quadratic x2 +2x4 1 x2 ... Leading term of a product is the product of leading terms
Read More At : www.math.utah.edu...

Suggested Questions And Answer :


Explain why the product (3x-3)(3x3) is a constant

?????????? "3x3" ???????? maebee that spozed tu be 3^3 ???? or maebee 3*3 ????? me dont see no konstant
Read More: ...

stating and explaining boyles law and charles law

It's all to do with proportion. Boyle's Law has the pressure of a gas inversely proportional to its volume. This means that as the volume of a fixed amount of gas decreases, its pressure increases, and vice versa, provided there is no change of temperature. This seems logical, because as a gas disperses (volume increases) the pressure it exerts drops. It also means that the product of pressure and volume is constant. Charles Law takes the volume and temperature as measurements, and keeps the pressure constant. This, too, is logical. If you heat a gas its volume increases and if you cool it the volume decreases. So volume is directly proportional to temperature. But there's a catch. The temperature isn't Fahrenheit or Celcius, because these are two different scales. Absolute temperature is what has to be measured which has its zero at absolute zero when the molecules of the gas are at rest. The temperature is in degrees Kelvin, which has is zero at -273 degrees Celcius. This means that all temperatures must be converted to degrees Kelvin, otherwise it has no meaning. So V is proportional to T where V=volume and T=absolute temperature.
Read More: ...

how to factorise fully

I can offer tips: Look out for constants and coefficients that are multiples of the same number, e.g., if all the coefficients are even, 2 is a factor. If 3 goes into all the coefficients, 3 is a factor. Place the common numerical factor outside brackets, containing the expression, having divided by the common factor. In 2x^2+10x-6 take out 2 to give 2(x^2+5x-3). In 24x^2-60x+36 12 is a common factor so this becomes: 12(2x^2-5x+3). This also applies to equations: 10x+6y=16 becomes 2(5x+3y)=2(8). In the case of an equation, the common factor can be completely removed: 5x+3y=8. Now look for single variable factors. For example, x^2y^3+x^3y^2. Look at x first. We have x^2 and x^3 and they have the common factor x^2, so we get x^2(y^3+xy^3) because x^2 times x is x^3. Now look at y. We have y^2 and y^3 so now the expression becomes x^2y^2(y+x). If the expression had been 2x^2y^3+6x^3y^2, we also have a coefficient with a common factor so we would get: 2x^2y^2(y+3x). If you break down the factors one at a time instead of all at once you won't get confused. After single factors like numbers and single variables we come to binomial factors (two components). These will usually consist of a variable and a constant or another variable, such as x-1, 2x+3, x-2y, etc. The two components are separated by plus or minus. In (2) above we had a binomial component y+x and y+3x. These are factors. It's not as easy to spot them but there are various tricks you can use to help you find them. More often than not you would be asked to factorise a quadratic expression, where the solution would be the product of two binomial factors. Let's start with two such factors and see what happens when we multiply them. Take (x-1) and (x+3). Multiplication gives x(x+3)-(x+3)=x^2+3x-x-3=x^2+2x-3. We can see that the middle term (the x term) is the result of 3-1, where 3 is the number on the second factor and 1 the number in the first factor. The constant term 3 is the result of multiplying the numbers on the factors. Let's pick a quadratic this time and work backwards to its factors; in other words factorise the quadratic. x^2+2x-48. The constant term 48 is the product of the numbers in the factors, and 2 is the difference between those numbers. Now, let's look at a different quadratic: x^2-11x+10. Again the constant 10 is the product of the numbers in the factors, but 11 is the sum of the numbers this time, not the difference. How do we know whether to use the sum or difference? We look at the sign of the constant. We have +10, so the plus tells us to add the numbers (in this case, 10+1). When the sign is minus we use the difference. So in the example of -48 we know that the product is 48 and the difference is 2. The two numbers we need are 6 and 8 because their product is 48 and difference is 2. It couldn't be 12 and 4, for example, because the difference is 8. What about the signs in the factors? We look at the sign of the middle term. If the sign in front of the constant is plus, then the signs in front of the numbers in the factors are either both positive or both negative. If the sign in front of the constant is negative then the sign in the middle term could be plus or minus, but we know that the signs within each factor are going to be different, one will be plus the other minus. The sign in the middle term tells us to use the same sign in front of the larger of the two numbers in the factors. So for x^2+2x-48, the numbers are 6 and 8 and the sign in front of the larger number 8 is the same as +2x, a plus. The factors are (x+8)(x-6). If it had been -2x the factors would have been (x-8)(x+6). Let's look at a more complicated quadratic: 6x^2+5x-21. (You may also see quadratics like 6x^2+5xy-21y^2, which is dealt with in the same way.) The way to approach this type of problem is to look at the factors of the first and last terms. Just take the numbers 6 and 21 and write down their factors as pairs of numbers: 6=(1,6), (2,3) and 21=(1,21), (3,7), (7,3), (21,1). Note that I haven't included (6,1) and (3,2) as pairs of factors for 6. You'll see why in a minute. Now we make a table (see (5) below). In this table the columns A, B, C and D are the factors of 6 (A times C) and 21 (B times D). The table contains all possible arrangements of factors. Column AD is the product of the "outside factors" in columns A and D and column BC is the product of the "inside factors" B and C. The last column depends on the sign in front of the constant 21. The "twiddles" symbol (~) means positive difference if the sign is minus, and the sum if the sign in front of the constant is plus. So in our example we have -21 so twiddles means the difference, not the sum. Therefore in the table we subtract the smaller number in the columns AD and BC from the larger and write the result in the twiddles column. Now we look at the coefficient of the middle term of the quadratic, which is 5 and we look down the twiddles column for 5. We can see it in row 6 of the figures: 2 3 3 7 are the values of A, B, C and D. If the number hadn't been there we've either missed some factors, or there aren't any (the factors may be irrational). We can now write the factors leaving out the operators that join the binomial operands: (2x 3)(3x 7). One of the signs will be plus and the other minus. Which one is which? The sign of the middle term on our example is plus. We look at the AD and BC numbers and associate the sign with the larger product. We are interested in the signs between A and B and C and D. In this case plus associates with AD because 14 is bigger than 9. The plus sign goes in front of the right-hand operand D and the minus sign in front of right-hand operand B. If the sign had been minus (-5x), minus would have gone in front of D and plus in front of B. So that's it: [(Ax-B)(Cx+D)=](2x-3)(3x+7). (The solution to 6x^2+5xy-21y^2 is similar: (2x-3y)(3x+7y).) [In cases where the coefficient of the middle term of the quadratic appears more than once, as in rows 1 and 7, where 15 is in the twiddles column, then it's correct to pick either of them, because it just means that one of the binomial factors can be factorised further as in (1) above.] Quadratic factors A B C D AD BC AD~BC 1 1 6 21 21 6 15 1 3 6 7 7 18 11 1 7 6 3 3 42 39 1 21 6 1 1 126 125 2 1 3 21 42 3 39 2 3 3 7 14 9 5 2 7 3 3 6 21 15 2 21 3 1 2 63 61  
Read More: ...

solve by factorisation the following equation (1).5x^2+2x-15=0 (2).21x^2-25x=4 (3).10x^2+3x-4=0

  (1) 5x^2+2x-15=0 I suspect that this should be 5x^2+22x-15=0, because the equation as written does not factorise. Write down the factors of the squared term coefficient and the constant term. We write these as ordered pairs (a,b) squared term (c,d): (1,5) and constant: (1,15), (15,1), (3,5), (5,3). The sign of the constant (minus) tells us that we are going to subtract the cross-products of  factors. If it had been plus we would be adding the cross-products. Now we create a little table: Quadratic factor table a b c d ad bc  |ad-bc| 1 5 15 1 1 75 74 1 5 1 15 15 5 10 1 5 3 5 5 15 10 1 5 5 3 3 25 22 The column |ad-bc| just means the difference between ad and bc, regardless of it being positive or negative, just take the smaller product away from the larger. If the coefficient of the x term is in the last column, then that row contains the factors you need. If it isn't in the last column, then the quadratic doesn't factorise, or you've missed some factors. If 22x is the middle term, then the factors are shown in the last row and (a,b,c,d)=(1,5,5,3). Now we look at the sign of the middle term. Whatever the sign is we put it in front of the number c or d for the larger cross-product. The cross-products are bc and ad. In this case, bc is bigger than ad so the + sign goes in front of c. The sign in front of the constant tells us whether the sign in front of d is different or the same. If the sign is plus it's the same, otherwise it's the opposite sign. So in this case, it's minus, so the minus sign goes in front of d and we have (ax+c)(bx-d) (note the order of the letters!) or (x+5)(5x-3), putting in the values for a, b, c and d. If (x+5)(5x-3)=0, then x+5=0 or 5x-3=0. So in the first case x=-5 and in the second case 5x=3 so x=3/5. (2) 21x^2-25x-4=0 (moving 4 over to the left to put the equation into standard form). Quadratic factor table a b c d ad bc  |ad-bc| 1 21 1 4 4 21 17 1 21 4 1 1 84 83 1 21 2 2 2 42 40 3 7 1 4 12 7 5 3 7 4 1 3 28 25 3 7 2 2 6 14 8 The row in bold print applies. The sign in front of 4 is minus so the signs in the brackets will be different. 28 is the larger product, so since we have -25x, the minus sign goes in front of c (4) and plus in front of d (1): (3x-4)(7x+1)=0. So the solution is x=4/3 or -1/7. (3) 10x^2+3x-4=0. Quadratic factor table a b c d ad bc  |ad-bc| 1 10 1 4 4 10 6 1 10 4 1 1 40 39 1 10 2 2 2 20 18 2 5 1 4 8 5 3 2 5 4 1 2 20 18 2 5 2 2 4 10 6 (2x-1)(5x+4)=0, so x=1/2 or -4/5.  
Read More: ...

How do special products help us factor polynomials? Give examples.

The special products that students are usually asked to identify and remember to help in factorisation are the squares and difference of squares: (a+b)^2=a^2+2ab+b^2; (a-b)^2=(b-a)^2=a^2-2ab+b^2; (a-b)(a+b)=a^2-b^2. a and b can be composite quantities like 3x, xy, 2xyz, etc. Examples: (3xy-z)^2=9x^2y^2-6xyz+z^2; (1/x+1/y)^2=1/x^2+2/xy+1/y^2; (5-t)(5+t)=25-t^2. I include some other types of products you might find useful: The constant term in a polynomial is a strong clue to how it factors, assuming that it is meant do so. If the constant is a prime number p, then the factors will consist of p and at least one 1, because p*1*...*1= p. The degree of the polynomial tells you how many factors there are: a degree 2 (quadratic) has two, for example. Examples: x^3-7x^2-x+7=(x-1)(x+1)(x-7); x^2-12x-13=(x-13)(x+1). If the constant is a composite number (i.e., not a prime number), then you need to factorise it and group the factors according to the degree. For example, if the constant is 15 and the degree is 4, then the factors arranged in fours are (1,1,1,15), (1,1,3,5). Example: x^4+2x^3-16x^2-2x+15=(x-1)(x+1)(x-3)(x+5). If the degree is n and the constant term is p^n, where p is a prime number, then the polynomial may consist of factors +p and -p. Examples: x^3-5x^2-25x+125=(x-5)^2(x+5). But x^2-26x+25=(x-25)(x-1). The sign of the constant is significant. If the degree n is even, and the sign is plus, then the zeroes of the polynomial will consist of an even number of pluses and minuses. If the sign is minus, there is an odd number of pluses and an odd number of minuses. If the degree is odd, and the sign is plus, the factors contain an even number of minuses, or none at all, and the rest are plus. If the sign is minus the factors contain an odd number of minuses, and the rest are pluses, or there are no other factors.
Read More: ...

find all real zeros of f(x)=6x^3-5x-1

Since 1 is the constant and the only roots of 1 have a magnitude of 1, we can see that ax=1 or ax=-1 is a root, where a is a constant. If we put x=1 into the function, it does in fact equal zero. So x-1 is a factor, which we can divide by, using synthetic division: 6x^2+6x+1. Therefore f(x)=(x-1)(6x^2+6x+1). The quadratic implies that we have to look at the factors of 6 (that is 6 and 1, and 2 and 3) and the factors of 1 (that is 1 and 1, and -1 and -1) and combine these possibilities to get the coefficient of the middle term, which is 6. We also know that from the sign of the constant we have to add the products of the prospective factors, and we know from that that we can have no minus signs. So we're looking for 1 and 1 as the constants, and the products are 2*1+3*1=5 or 6*1+1*1=7. Neither of these gives us the 5 we need. The solution of the quadratic is x=(-6+sqrt(36-24))/12=(-3+sqrt(3))/6=-0.2113 or -0.7887. Therefore the three roots are 1, -0.2113, -0.7887 approx.
Read More: ...

How do you factor the sum of terms as a product of the GCF and a sum?

It took me a while to interpret your question, and I came up with a polynomial that had a GCF, but it doesn't have to be a polynomial: 36x^2+45x+63 simplifies when we note that the GCF of the numbers 36, 45 and 63 is 9. That means 9 is the largest integer to divide into the numbers: 36=9*4, 45=9*5 and 63=9*7. This means we can rewrite the polynomial as the product of a GCF and a sum: 9(4x^2+5x+7). It utilises the distributive property of numbers. Another example: abc+ace-ca^2. In this case the GCF is algebraic: ac, because all the terms contain ac, so we can factorise: ac(b+e-a). This is the product of the GCF ac and the sum b+e-a. Another example: 4x^2+2xy+6x=2x(2x+y+3). This time the GCF is 2x because the GCF of the numbers is 2 and of the algebraic quantities is x, so we just combine the two GCFs to make 2x. Does this help you in time? If you still don't understand send me a private message explaining your difficulties and providing examples.
Read More: ...

If y varies directly with x, does x vary directly with y?

If y varies directly with x, does x vary directly with y? If so, what is the relationship between the constants of variation? Explain. Please help! x is the constant y is the variable. the answer is no. y is dependent on x and therefore does not vary.
Read More: ...

6x²+13xy+11x+36y-5y²-7

This is the expansion of (ax+by+c)(dx+ey+f) where a to f represent numerical constants (integers), obeying the following arithmetic rules: ad=6, be=-5, cf=-7 (coefficients of x^2, y^2, constant) The only numbers we have for a, b, c, d, e or f must be in the set {1 1 1 2 3 5 6 7}; no other numbers can be used and the arithmetic rules must always apply. This is because the only factors of 5 are 1 and 5; the only factors of 7 are 1 and 7; but the pairs of factors of 6 are 1 and 6 or 2 and 3. I haven't yet considered the signs plus and minus. We also have the sums of two products: ae+bd=13 (coefficient of xy), af+cd=11 (x), ce+bf=36 (y). The largest of these is 36, and there's only one way of using 4 numbers in the set to get 36 using each one once and that is 7*5+1*1. So from this we can find b, c, e and f. The following table shows all 8 combinations for these coefficients: bf+ce Ref b c e f 1 1 -7 -5 1 2 1 -5 -7 1 3 -7 1 1 -5 4 -5 1 1 -7 5 -1 7 5 -1 6 -1 5 7 -1 7 7 -1 -1 5 8 5 -1 -1 7 The remaining two coefficients are a and d and the possible pairs of factors are 1 and 6, or 2 and 3. Here's a table to show all 8 possible combinations: a and d Ref a d 1 1 6 2 6 1 3 -1 -6 4 -6 -1 5 2 3 6 3 2 7 -2 -3 8 -3 -2 The next bit is tedious. We need to try out all the combinations of b, c, e and f against all those of a and d! Yes, that's 64 in all. What we're looking for is the sum of the products ae+bd=13. I'll spare you the details and just show you the results. Ref 1 from the first table with ref 8 from the second gives the required result and also satisfies af+cd=11. So we have the values of a to f: -3, 1, -7, -2, -5, 1, and the factorisation is: (-3x+y-7)(-2x-5y+1). This can also be written: (3x-y+7)(2x+5y-1), which is ref 5 from the first table with ref 6 from the second. Combinations in red have complementary signs to those shown in black.  
Read More: ...

Write an inequality that represents the fact that while making each product you can't exceed the amount invested ($5,000) for a business where the products will cost the owner $2.50 and $3.25 to make.

A. 5000/2.5=2000 and 5000/3.25=1538.5. If he only makes $2.50 items he can't exceed 2000, and if he only makes $3.25 items he can't exceed 1538. So the inequality is 2.5A+3.25B<5000 where A represents the number of $2.50 items and B that of $3.25 items. B. To graph the inequality, we write one item in terms of the other: 3.25B=5000-2.5A. If B is the y axis and A the x axis we get a backward sloping straight line. The y intercept is 5000/3.25=1538.5 and the x intercept is 5000/2.5=2000. Choose a suitable scaling and join the intercepts. The graph only has a positive region so the origin of the axes can be placed very much on the left and towards the bottom of the page. The shaded area is the region bounded by the line and axes. C. Choose the three points P, Q and R as directed. P is within the shaded area and represents (A,B) in budget; Q makes use of the whole investment; R is outside budget (exceeds the investment). Although the graph is a straight line, in reality it is a sequence of points because A and B cannot contain fractions. If the graph has been drawn reasonably accurately it should be possible to read off values of A and B. The values A and B must be integers, so they need to be rounded down to the nearest integer. To work out the cost calculate C=2.5A+3.25B. The quantity Q=A+B. The average unit cost=C/Q.  
Read More: ...

Tips for a great answer:

- Provide details, support with references or personal experience .
- If you need clarification, ask it in the comment box .
- It's 100% free, no registration required.
next Question || Previos Question
  • Start your question with What, Why, How, When, etc. and end with a "?"
  • Be clear and specific
  • Use proper spelling and grammar
all rights reserved to the respective owners || www.math-problems-solved.com || Terms of Use || Contact || Privacy Policy
Load time: 0.0402 seconds