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Simplify 3 · 2x. What is the coefficient?

Find the coefficient

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Do the operation 3*2=6. Put the 6 in front of x. 6 is the coefficient
The result, in simplified form, is 6x

Posted on Mar 28, 2014

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1.2(1.5x-2.2y)-4.5(2.0x+1.5y)


1.2(1.5x) + 1.2(-2.2y) + -4.5(2x) + -4.5(1.5y)

Figure out the products (multiplication), then simplify by adding or subtracting the coefficients of the same variable ( the X and the Y).


coefficients Google Search

May 30, 2016 | Office Equipment & Supplies

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A linear equation is given below. Determine the steps to solve the equation for its inverse. y=6x+24


Step 1 - replace each x with a y and replace each y with an x
x=6y+24
Step 2 - Solve for y to get the equation in the slope intercept form y=mx+b
Step 3 - get the 24 to the other side by subtracting 24 for both sides
x-24=6y + 24 - 24
Simplify
x - 24 = 6y
Divide by the coefficient in front of the y to get y by itself.
x/6 - 24/6 = y
x/6 -4 = y

Write in the normal format by switching the sides.

y = x/6 - 4

Good luck.

Paul

May 07, 2016 | Office Equipment & Supplies

1 Answer

Chacha answers


It seems to me that you are trying to solve the quadratic equation
aX^2+bX+c=10 with a=-3, b=3, c=15 or -3X^2+3X+15=0.
Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation -3X^2+3X+15=0..
You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).
Discriminant is usually represented by the Greek letter DELTA (a triangle)
DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189
If the discriminant is positive (your case) the equation has two real solutions which are given by
Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2
Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2 or about -1.791287847
Here SQRT stands for square root.

Aug 17, 2014 | Computers & Internet

1 Answer

Chacha answers


It seems to me that you are trying to solve the quadratic equation
aX^2+bX+c=10 with a=-3, b=3, c=15 or -3X^2+3X+15=0.
Since the all the coefficients are multiples of 3, one can simplify the equation by dividing every thing by 3, leaving -X^2+X+5=0. But to avoid confusing you I will consider the original equation -3X^2+3X+15=0..
You must first find out if the equation has any real solutions. To do that you calculate the discriminant (you do not have to remember the name if you choose to).
Discriminant is usually represented by the Greek letter DELTA (a triangle)
DELTA =b^2-4*a*c =(3)^2-4*(-3)*(15)=189
If the discriminant is positive (your case) the equation has two real solutions which are given by
Solution1 =X_1=(-3-SQRT(189))/(-2*3)=(1+SQRT(21))/2
Solution2 =X_2=(-3+SQRT(189))/(-2*3)=(1-SQRT(21))/2 or about -1.791287847
Here SQRT stands for square root.

Aug 17, 2014 | SoftMath Algebrator - Algebra Homework...

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Ti-89 titanium diagnostics


There is no DiagnosticsOn command in the TI 89 Titanium. The command exists on the TI83/84PLUS where it displays the r and r^2 coefficients. On the TI89 , the correlation coefficient and the determination coefficient (r, r^2) are displayed by default among the statistics if THE REGRESSION MODEL includes them. If they are not defined by the regression model they will not be displayed. In short, you do not have to turn diagnostics ON.

Apr 17, 2012 | Texas Instruments TI-89 Calculator

1 Answer

Casio FX - 9860G SD have complex equations??


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.

Mar 16, 2012 | Casio Office Equipment & Supplies

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

2 Answers

Factoring polynomials "Algebra"


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

Nov 18, 2007 | Casio Office Equipment & Supplies

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