Question about XRS-757 Radar Detector

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Here is a great explanation of radar types

http://www.radarbusters.com/detectorsfaqsarticle.cfm

Posted on Aug 04, 2010

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Posted on Jan 02, 2017

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The calculator cannot do symbolic algebra. If equation is aX^2+bX+c=0, write it in the form a(X^2+(b/a)X+(c/a))=0 and solve the quadratic equation X^2+(b/a)X+(c/a)=0. Get the approximate roots X1, and X2 (if they exist) and write your original equation in the for

a(X-X1)(X-X2)=0

Your quadratic polynomial is factored as** a(X-X1)(X-X2)**

a(X-X1)(X-X2)=0

Your quadratic polynomial is factored as

May 28, 2014 | Casio Office Equipment & Supplies

ALL VISION BATTERY MODELS

Date code stamped on top of battery container

X1 X2 X3 X4

Where X1 denotes year of manufacture A = 2001 B = 2002 etc.

X2 denotes month of manufacture A = January B = February etc.

X3 X4 denotes date of month of manufacture

Eg AH27 = battery manufacturing date 27th August, 2001.

Date code stamped on top of battery container

X1 X2 X3 X4

Where X1 denotes year of manufacture A = 2001 B = 2002 etc.

X2 denotes month of manufacture A = January B = February etc.

X3 X4 denotes date of month of manufacture

Eg AH27 = battery manufacturing date 27th August, 2001.

Dec 12, 2013 | Exide Chloride Power Station (EXP2000) UPS...

The simultaneous equation solver requires the coefficients to be real. Similarly matrices must have real coefficients.

Your only alternative is to express each of A and B as a real part and imaginary part.

A= x1+iy1

B=x2+iy2.

Substitute x1+iy1 for A in the two equations. Substitute x2+iy2 for B in the two equations. Do the algebra. Gather real parts and gather imaginary parts. Split each original equation into two equations: One equation comes from setting Real Part of left side = real part of right side (1); the other equations comes from setting the imaginary part of left side= imaginary part of right side (here 0).

Do the same procedure for the 2nd original equation.

At the end of the process you will have 4 coupled linear equations in the 4 unknowns (x1,y1,x2,y2).

Then you might want to use the calculator to solve the derived system. Once you have x1,y1,x2,y2 you reconstruct A=x1+iy1, etc.

Your only alternative is to express each of A and B as a real part and imaginary part.

A= x1+iy1

B=x2+iy2.

Substitute x1+iy1 for A in the two equations. Substitute x2+iy2 for B in the two equations. Do the algebra. Gather real parts and gather imaginary parts. Split each original equation into two equations: One equation comes from setting Real Part of left side = real part of right side (1); the other equations comes from setting the imaginary part of left side= imaginary part of right side (here 0).

Do the same procedure for the 2nd original equation.

At the end of the process you will have 4 coupled linear equations in the 4 unknowns (x1,y1,x2,y2).

Then you might want to use the calculator to solve the derived system. Once you have x1,y1,x2,y2 you reconstruct A=x1+iy1, etc.

Aug 29, 2011 | Casio FX-9860G Graphic Calculator

Ok its one minute time cause x1=y+2 x2=y+1 x3=x1-x2 where y is the column speed

Jul 19, 2011 | Computers & Internet

The site seems to eat the plus signs I enter, so I will use PLUS to symbolize addition.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

Jan 27, 2011 | Texas Instruments TI-84 Plus Calculator

Hello,

The Casio FX-300ES does not handle symbolic algebra. So it cannot factor a general polynomial expression. The methods can be found in any book on Algebra.

However if you are interested in approximate factorization of quadratic and cubic polynomials, you can use the calculator to do that. It can solve aX^3 +bX^2+cX+d =0 and the quadratic equations.

If you want to factor a cubic polynomial P3(X) = aX^3+bX^2+cX+d , you write the corresponding cubic equation as**aX^3+bX^2+cX=d =0** , then you divide all terms of the equation by** a** to obtain

**X^3+(b/a)X^2+(c/a)X+(d/a)=0.**

You use the calculator to solve (approximately) this equation.

Suppose you find the 3 roots**X1,X2,and X3.** Then the polynomial X^3+(b/a)X^2+(c/a)X+(d/a) can be cast in the factored form (X-X1)(X-X2)(X-X3) and the original polynomial P3(X) can be written as

**P3(X) = a*(X-X1)(X-X2)(X-X3) **

You can handle the quadratic polynomial the same way.

P2(X) =a*(X-X1)(X-X2) where X1, X2 are the two real roots

Hope it helps.

The Casio FX-300ES does not handle symbolic algebra. So it cannot factor a general polynomial expression. The methods can be found in any book on Algebra.

However if you are interested in approximate factorization of quadratic and cubic polynomials, you can use the calculator to do that. It can solve aX^3 +bX^2+cX+d =0 and the quadratic equations.

If you want to factor a cubic polynomial P3(X) = aX^3+bX^2+cX+d , you write the corresponding cubic equation as

You use the calculator to solve (approximately) this equation.

Suppose you find the 3 roots

You can handle the quadratic polynomial the same way.

P2(X) =a*(X-X1)(X-X2) where X1, X2 are the two real roots

Hope it helps.

Sep 27, 2009 | Casio fx-300ES Calculator

Hello,

This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.

Let the equation be ax^2 +bx+c=0

You compute the discriminant Call it disc.

**disc= (b^2-4*a*c)**

1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.

Then** x1=(-b+square root of disc)/(2*a)**

and**x2= (-b-square root of disc)/(2*a)**

2.** If disc=0** , square root of disc =0 and** x1=x2=-b/(2*a)**

3.**If disc is negative, there are no real solutions.**

To find the solutions you**replace a, b, and c **by their numerical values, to calculate disc, its square root, and calculate X1 and X2.

Hope it helps.

This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.

Let the equation be ax^2 +bx+c=0

You compute the discriminant Call it disc.

1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.

Then

and

2.

3.

To find the solutions you

Hope it helps.

May 05, 2009 | Sharp EL-531VB Calculator

Hello,

Sorry, but you cannot use this calculator to factorize a general polynomial.

1. It does not know symbolic algebra.

2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients (no letters) you can set [MODE] to equation and use the equation solver to find the real roots of 2nd degree or 3rd degree polynomials. Assuming that your polynomials have real roots (X1, X2) for the polynomial of degree 2, or (X1, X2, X3) for the polynomial of degree 3, then it is possible to write

P2(X) =a*(X-X1)(X-X2)

P3(X)= a(X-X1)(X-X2)(X-X3)

This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)

where a is the coefficient of the highest degree monomial aX^2 +...

or aX^3 +....

But I have a hunch that this is not what you wanted to hear.

Good luck.

Sorry, but you cannot use this calculator to factorize a general polynomial.

1. It does not know symbolic algebra.

2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients (no letters) you can set [MODE] to equation and use the equation solver to find the real roots of 2nd degree or 3rd degree polynomials. Assuming that your polynomials have real roots (X1, X2) for the polynomial of degree 2, or (X1, X2, X3) for the polynomial of degree 3, then it is possible to write

P2(X) =a*(X-X1)(X-X2)

P3(X)= a(X-X1)(X-X2)(X-X3)

This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)

where a is the coefficient of the highest degree monomial aX^2 +...

or aX^3 +....

But I have a hunch that this is not what you wanted to hear.

Good luck.

Mar 08, 2009 | Casio fx-300ES Calculator

Hello,

Sorry, but you cannot use this calculator to factor a general polynomial.

1. It does not know symbolic algebra.

2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients**
(no letters) **you can set [MODE] to **Equation **and use the equation solver
to find the real roots of 2nd degree or 3rd degree polynomials.
Assuming that your polynomials have real roots (X1, X2) for the
polynomial of degree 2, or (X1, X2, X3) for the polynomial of degree 3,
then it is possible to write

**P2(X) =a*(X-X1)(X-X2)**

P3(X)= a(X-X1)(X-X2)(X-X3)

where a is the coefficient of the highest degree monomial aX^2 +...

or aX^3 +....

This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)

While the [MODE][5:Equation] only handles quadratic and cubic equations, the [SHIFT][SOLVE=] solver finds the roots of arbitarry expressions (not limited to polynomials). In principle you can use it to find the roots of an expression. If it is a polynomial of dgree higher that 3 you can factor it (approximately).

But I have a hunch that this is not what you wanted to hear.

Hope it helps.

Sorry, but you cannot use this calculator to factor a general polynomial.

1. It does not know symbolic algebra.

2. It can only manipulate numbers.

However if you have polynomials of degree 2 or 3, with numerical coefficients

P3(X)= a(X-X1)(X-X2)(X-X3)

where a is the coefficient of the highest degree monomial aX^2 +...

or aX^3 +....

This is an approximate factorization, except if your calculator configured in MathIO, has been able to find exact roots (fractions and radicals)

While the [MODE][5:Equation] only handles quadratic and cubic equations, the [SHIFT][SOLVE=] solver finds the roots of arbitarry expressions (not limited to polynomials). In principle you can use it to find the roots of an expression. If it is a polynomial of dgree higher that 3 you can factor it (approximately).

But I have a hunch that this is not what you wanted to hear.

Hope it helps.

Dec 09, 2008 | Casio fx-300ES Calculator

X stands for wide band radars(easy to detect,lots of false alarms)

K/Ka stands for pulse band radars(similar signal has auto doors)

W,L stands for laser detection (very little false signals)

Ku stands for signal most used on speed cameras

V stands for radar detectors detector

Have in mind that you can only lower your speed 20-30 km/h in city zones, and up to 50 km/h on highway.This depends on car,brakes and You.

K/Ka stands for pulse band radars(similar signal has auto doors)

W,L stands for laser detection (very little false signals)

Ku stands for signal most used on speed cameras

V stands for radar detectors detector

Have in mind that you can only lower your speed 20-30 km/h in city zones, and up to 50 km/h on highway.This depends on car,brakes and You.

Jun 16, 2008 | XRS-757 Radar Detector

Jan 31, 2014 | XRS-757 Radar Detector

Nov 29, 2013 | XRS-757 Radar Detector

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