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# 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 ;

### 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

### 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

### 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 ...

### 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

### 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.

### 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

### 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

## 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

### 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.

### 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.

### 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.

### 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.

### 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.

### 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.

### 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.