Question about Toshiba Satellite A105 Notebook

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2x+7y-16=0

Posted on Aug 16, 2008

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The three x^2 coeficients are 3 -6 -2. Add them together. Answer -5.

Posted on Jun 13, 2008

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Mather chod

Posted on Mar 08, 2008

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Posted on Jan 02, 2017

Do the operation 3*2=6. Put the 6 in front of x. **6 is the coefficient**

The result, in simplified form, is 6x

The result, in simplified form, is 6x

Mar 26, 2014 | Office Equipment & Supplies

The Casio FX-9860G SD can solve a polynomial equation
of degree 2 or 3 with REAL coefficients. If the complex MODE is set to
REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.

The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).

Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Mar 17, 2012 | Casio Office Equipment & Supplies

Your question (?) is not clear so I will not try to guess what you really mean.

Be it as it may, the Casio FX-9860G SD can solve a polynomial equation of degree 2 or 3 with REAL coefficients. If the complex MODE is set to REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Be it as it may, the Casio FX-9860G SD can solve a polynomial equation of degree 2 or 3 with REAL coefficients. If the complex MODE is set to REAL it will find the real roots. If the complex mode is set to

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Mar 16, 2012 | Casio Office Equipment & Supplies

Your question (?) is not clear so I will not try to guess what you really mean.

Be it as it may, the Casio FX-9860G SD can solve a polynomial equation of degree 2 or 3 with REAL coefficients. If the complex MODE is set to REAL it will find the real roots. If the complex mode is set to** a+ib**, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Be it as it may, the Casio FX-9860G SD can solve a polynomial equation of degree 2 or 3 with REAL coefficients. If the complex MODE is set to REAL it will find the real roots. If the complex mode is set to

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Mar 16, 2012 | Casio FX-9750GPlus Calculator

y=x2-4x+3

here,a=1(coefficient of x2)

b=-4(coefficient of x)

c=3(the constant term)

as this is a quadratic equation this will have two solutions

solution 1=(-b+square root of(b2-4ac))/2a

=(-(-4)+square root of(16-4*a*c)/2*1

=3

solution 2=(-b-square root of(b2-4ac))/2a

=1

here,a=1(coefficient of x2)

b=-4(coefficient of x)

c=3(the constant term)

as this is a quadratic equation this will have two solutions

solution 1=(-b+square root of(b2-4ac))/2a

=(-(-4)+square root of(16-4*a*c)/2*1

=3

solution 2=(-b-square root of(b2-4ac))/2a

=1

Aug 07, 2011 | TOPICS Entertainment Snap! Geometry...

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:**

**Now to solve using the Ti 89.**

Using

(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

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

**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. **

Jan 04, 2011 | Texas Instruments TI-89 Calculator

I assume you want to solve a linear system of three equations i three unknowns (x,y,z).

- Turn calculator ON.
- If you do not see icons, press [MENU] key.
- Use arrows to move focus to [EQUA] and press [EXE]
- In new screen select F1: Simultaneous
- Select the numbewr of unknows (in your case 3): Press [F2:3]
- A table opens: it has 3 rows and 4 columns
- Enter the coefficients in first equation on first line (a1, b1,c1) and the constant term d1.
- Enter the other coefficients and the constant terms at the right places, pressing [EXE] after each entry.
- When finished coefficients and constant terms, press [F1:Solv]
- The solution vector (X,Y,Z) is displayed on next screen.
- Note: if the coefficients are integers, it is possible that the solution vector will be displayed as mixed fractions (exact solutions),
- If not, the solution vector will be in decimal format.

Mar 02, 2010 | Casio FX-9750GPlus Calculator

I assume you are speaking of solving a system of equations with a number of unknowns. If not, please correct me. Here's an example in practice:

If you have a system of 3 equations with 3 unknowns, you would set up your matrix so that the coefficients of each variable for a particular equation are on one row. So, given equations x + y + z = 0, 2x + 3y - 4z = 1, x + -z = -1 you would type the following into your calculator: [[1,1,1,0][2,3,-4,1][1,0,-1,-1]] and press enter to make sure you typed it correctly. notice that in the third row there is a zero, since we have zero time y for the third equation. Then row-reduce the matrix (2nd > 5 > 4 > 4 or in the CATALOG as rref). You should get out the matrix [[1,0,0,-1][0,1,0,1][0,0,1,0]]. This says that x=-1 y = 1 z=0 since my first column contained the coefficients for the x variable, the second for the y variable, and the third for the z variable. The last column contains the solution, the part on the other side of the equals sign.

Hope this helps! For more reading (from someone else; I just made this one up), check out the Wikipedia articles on Gaussian elimination and Systems of linear equations

If you have a system of 3 equations with 3 unknowns, you would set up your matrix so that the coefficients of each variable for a particular equation are on one row. So, given equations x + y + z = 0, 2x + 3y - 4z = 1, x + -z = -1 you would type the following into your calculator: [[1,1,1,0][2,3,-4,1][1,0,-1,-1]] and press enter to make sure you typed it correctly. notice that in the third row there is a zero, since we have zero time y for the third equation. Then row-reduce the matrix (2nd > 5 > 4 > 4 or in the CATALOG as rref). You should get out the matrix [[1,0,0,-1][0,1,0,1][0,0,1,0]]. This says that x=-1 y = 1 z=0 since my first column contained the coefficients for the x variable, the second for the y variable, and the third for the z variable. The last column contains the solution, the part on the other side of the equals sign.

Hope this helps! For more reading (from someone else; I just made this one up), check out the Wikipedia articles on Gaussian elimination and Systems of linear equations

May 03, 2009 | Texas Instruments TI-84 Plus Calculator

Mode>choose "5"

choose equation format "4"

input the coefficient for a which is "1" and hit "="

input the coefficient for b which is "4" and hit "="

input the coefficient for c which is "3" and hit "="

input the coefficient for d which is "12" and hit "="

Hit "=" for X1

Hit "=" again for X2

Hit "=" again for X3

choose equation format "4"

input the coefficient for a which is "1" and hit "="

input the coefficient for b which is "4" and hit "="

input the coefficient for c which is "3" and hit "="

input the coefficient for d which is "12" and hit "="

Hit "=" for X1

Hit "=" again for X2

Hit "=" again for X3

Nov 18, 2007 | Casio Office Equipment & Supplies

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