Question about Texas Instruments TI-89 Calculator

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{Use the binomial theorem to find the eight term of (3x-2y)^13}

The Choice are. (-160123392x^6y^7) (120092544x^7y^8) (-120092544x^7y^8) (None of the preceding)

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This may help:
http://en.wikipedia.org/wiki/Binomial_theorem
Rate me, thanks.

Posted on Jul 08, 2009

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What is common denominator of this equation? 10/2y+8. 7y+8/y2-16. -8/2y-8=


Recall that y^2-16=(y+8)*(y-8). That is the common denominator of the three rational expressions .

Oct 08, 2014 | Office Equipment & Supplies

1 Answer

2x^3-3x (a) state whether the polynomial is a monomial,binomial, trinomial, or none of these. (b) state whether it is in descending or ascending order. provide a reasons for your answer


a) It is a 3rd degree binomial because there are only 2 terms.

b) It is in descending order because the degree of each term goes down as you read the expression from left to right

May 23, 2014 | Office Equipment & Supplies

1 Answer

Demonstrate how to multiply two binomials


Try to seek these: By using the suggested format:
Suggested su
  1. Adding and Subtracting Binomials
    • 1

      Arrange each term in each binomial in order of degree from greatest to least. The degree of a binomial is the exponent attached to the term. For example, 4x^2 is a second degree term.

    • 2

      Multiply each term in the binomial that is being subtracted by -1 to turn it into an addition problem. For example, the problem (8x^2 + 8) - (x^2 - 2) becomes (8x^2 + 8) + (-x^2 + 2).

    • 3

      Combine like terms. In the example problem, the x^2 terms are combined and the constant terms are combined, yielding (8x^2 + 8) + (-x^2 + 2) = 7x^2 + 10.

    Multiplying Binomials
    • 4

      Understand the F.O.I.L. method. F.O.I.L. is an acronym standing for first, outside, inside and last. It means that you multiply the first number of the first binomial by the first number of the second, then the numbers on the outside (the first term of the first binomial by the second term of the second binomial) and so on. This ensures that both numbers in the first binomial are multiplied by both numbers in the second.

    • 5

      Use the F.O.I.L. method to multiply the two binomials together. For example, (3x + 4)(3x - 4) = 9x^2 +12x - 12x - 16. Notice that -12x is the product of the outside terms and -16 is the product of the last terms, 4 and -4.

    • 6

      Simplify. There will almost always be like terms to combine. In the example, 12x and -12x cancel out, yielding the answer 9x^2 - 16.

    Dividing Binomials
    • 7

      Use the distributive property to divide both terms in the binomial by the monomial divisor. For example, (18x^3 + 9x^2) / 3x = (18x^3 / 3x) + (9x^2 / 3x).

    • 8

      Understand how to divide by a term. If you are dividing a higher order term by a lower order term, you subtract the exponent. For example, y^3/y = y^2. The number part of each term is handled like any other division problem. For example, 20z / 4 = 5z.

    • 9

      Divide each term in the binomial by the divisor; (18x^3 / 3x) + (9x^2 / 3x) = 6x^2 + 3x.

May 10, 2014 | Computers & Internet

1 Answer

(3x-2y)^4 use the binomial theorem


You have to type expand((38x-28y)^4,x). See captured images
2_9_2012_10_43_20_am.jpg

Jan 04, 2012 | Texas Instruments TI-89 Calculator

1 Answer

Solve for x and y: 0.2x+0.7y=1.5 0.3x-0.2y=1


Ok so first you need to find what x equals in terms of y. so use the first equation .2x+.7y=1.5 solve for x by the following steps
subract .7y from both sides: .2x= 1.5-.7y --->
divide by .2 on both sides: x= 7.5 - 3.5y. --->
Now plug in x= 7.5 - 3.5y for x in the second equation 0.3 (7.5 - 3.5y) -0.2y=1 --->
Distribute the .3: 2.25 - 1.05y -.2y = 1--->
solve for y by combining like terms (the y's) and subtractive 2.25: -1.25y = -1.25--->
divide by -1.25 and you get y= 1
Now go back to your x= 7.5 - 3.5y you solved for. Plug in 1 for y in the equation and solve for x--->
x=7.5-3.5(1) so x= 4 and y=1 yay your done!

May 03, 2011 | Texas Instruments TI-30 XIIS Calculator

2 Answers

TI-89 Titanium: I want to solve a Binomial Theorem problem (x+y)^6 how would i go about solving this in the calculator?


Using elementary algebria in the binomial theorem, I expanded the power (x + y)^n into a sum involving terms in the form a x^b y^c. The coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n. This is known as binomial coefficients and are none other than combinatorial numbers.

Combinatorial interpretation:

Using binomial coefficient (n over k) allowed me to choose k elements from an n-element set. This you will see in my calculations on my Ti 89. This also allowed me to use (x+y)^n to rewrite as a product. Then I was able to combine like terms to solve for the solution as shown below.
(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y) = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

This also follows Newton's generalized binomial theorem:


oneplusgh_15.jpg
Now to solve using the Ti 89.


Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:

oneplusgh.gif

The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n ? k) + k, always equals n.



oneplusgh_26.jpg


The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.



oneplusgh_18.jpg

This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.


oneplusgh_19.jpg

This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y



oneplusgh_20.jpg


This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2



oneplusgh_27.jpg



This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3



oneplusgh_28.jpg



This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4



oneplusgh_21.jpg


This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5



oneplusgh_22.jpg

This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6



oneplusgh_23.jpg

Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.

binomial coefficient (n over k) for (x+y)^6
x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6












Jan 04, 2011 | Texas Instruments TI-89 Calculator

1 Answer

(2x+3y)(3x+2y)


(2x - 3y)(3x + 2y) = 2x*3x + 2x*2y - 3y*3x - 3y*2y
= 6x^2 + 4xy - 9xy - 6y^2
= 6x^2 - 5xy - 6y^2 You multiply each element in the first set of brackets by each element in the second set of brackets and then consolidate like terms and arrange them in sequence of powers, first x and then y. So:
2x by 3x = 6x squared
2x by 2y = 4xy
-3y by 3x = -9xy
-3y by 2y = 6y squared

6x squared -5xy + 6y squared

Dec 04, 2010 | Yahoo Computers & Internet

3 Answers

Solving systems of equations by elimination


Hi joanmae jmeh;

You can add these equations just like you would numbers

3x + 2y = 7
5x - 2y = 1
-----------------------
8x =8
x=1
Now plug x = 1 back into either equation and you can solve for y
Y = 2

When you add the 2 equations together the +2y and the -2y cancel
out

Hope this helps Loringh Please leave a rating for me Thks

Dec 01, 2008 | Bagatrix Algebra Solved! 2005 (105101) for...

2 Answers

What type of problem is this (9x-7)-4(3x 5)


just a simple multiplication solve first which is inside the brackets 9x(-7)=-63 3x5=15 (-63)-4x15 -67x15 -1005 .. i think am rite..

Oct 12, 2008 | SoftMath Algebrator - Algebra Homework...

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