Question about Vivendi Excel@ Mathematics Study Skills (71101) for PC, Mac

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Posted on Jan 02, 2017

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This is best written as two separate equations:

8x+3y = -23 and 34x+ 5y = -23

Solving the first one for x:

8x = -23-3y

x = -23/8 - 3/8y

Substituting this value for x into the second equation:

34(-23/8 - 3/8y) + 5y = -23

-97.75 - (34)(.375)y + 5y = -23

-97.75 - 12.75y + 5y = -23

-97.75 -7.75y = -23

-7.75y = 97.75-23=74.75

**y **= -74.75/7.75 =** -9.645161**

Substitution back into the equation for x:

x = -23/8 - 3/8(-9.645161)

x = -2.875 + 3.616935

**x** **=.741935**

8x+3y = -23 and 34x+ 5y = -23

Solving the first one for x:

8x = -23-3y

x = -23/8 - 3/8y

Substituting this value for x into the second equation:

34(-23/8 - 3/8y) + 5y = -23

-97.75 - (34)(.375)y + 5y = -23

-97.75 - 12.75y + 5y = -23

-97.75 -7.75y = -23

-7.75y = 97.75-23=74.75

Substitution back into the equation for x:

x = -23/8 - 3/8(-9.645161)

x = -2.875 + 3.616935

Dec 12, 2014 | Bagatrix Algebra Solved! 2005 (105101) for...

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

First of all you equation is not one : it has nothing on the right side of the = sign. But to answer the general question let us write the equation as **50+25x-5y=0**

**X-intercept (also know as roots) There may be several**

Definition: X-intercepts are those values of the independent variable x**for which y=0**. For a straight line there con be at most 1 x-intercept.

To find the intercept, set y=0 in the equation of the line and solve for x

50+25x-5(0)=0 or 50+25x=0. The solution is** x=-(50/25)=-2**

**Y-intercept (also know as the initial value.** There can only be 1 y-intercept, otherwise the expression does not represent a function.

Definition: It is the value of the dependent variable y when x=0 (where the function crosses the y-axis

To find it, set the x-value to 0 in the equation of the line.

**50+25x-5y=0**

50+25(0)-5y=0, or 50-5y=0. The solution is**y=50/5=10**

The straight line cuts the x-axis at the point (-2, 0) and the y-axis at the point (0,10)

Definition: X-intercepts are those values of the independent variable x

To find the intercept, set y=0 in the equation of the line and solve for x

50+25x-5(0)=0 or 50+25x=0. The solution is

Definition: It is the value of the dependent variable y when x=0 (where the function crosses the y-axis

To find it, set the x-value to 0 in the equation of the line.

50+25(0)-5y=0, or 50-5y=0. The solution is

The straight line cuts the x-axis at the point (-2, 0) and the y-axis at the point (0,10)

Jan 28, 2014 | Computers & Internet

Calcualte the slope of the line as

a=(7-(-11))/(10-(-5))=18/15=6/5

Use the fact that the line passes through one of the two points, for example (10,7)

7=(6/5)*10+b=12+b

Obtain b as b=7-12=-5

The equation of the line in functional form is y=(6/5)x-5

Multiply everything by 5 to clear the fraction

5y=6x-25 or 0=6x-5y-25

Finally, the equation in general form (standard?) is**6x-5y-25=0**.

Check the calculation by verifying that the point (10,7) lies on the line.

6(10)-5(7)-25=60-35-25=60-60=0 CHECKed!

Check that the second point (-5,-11) lies on the line also (if you want to)

6*(-5)-5*(-11)-25=-30+55-25=0

That checks OK.

a=(7-(-11))/(10-(-5))=18/15=6/5

Use the fact that the line passes through one of the two points, for example (10,7)

7=(6/5)*10+b=12+b

Obtain b as b=7-12=-5

The equation of the line in functional form is y=(6/5)x-5

Multiply everything by 5 to clear the fraction

5y=6x-25 or 0=6x-5y-25

Finally, the equation in general form (standard?) is

Check the calculation by verifying that the point (10,7) lies on the line.

6(10)-5(7)-25=60-35-25=60-60=0 CHECKed!

Check that the second point (-5,-11) lies on the line also (if you want to)

6*(-5)-5*(-11)-25=-30+55-25=0

That checks OK.

Dec 04, 2011 | Super Tutor Pre Algebra (ESDPALG)

x-intercept (or zero of the function)

Set y=0 in the equation and solve fo x: 2x-5(0)=20 and x=20/2=10

y-intercept (initial value)

Set x= 0 in the equation and solve for y: 2(0)-5y=20 and y=20/(-5)=-4

If the equation is in the slope-intercept form y=ax+b

1. y-intercept is b

2. To get the x-intercept ( zero of the function) Set y=0=ax+b, and x=-b/a

Set y=0 in the equation and solve fo x: 2x-5(0)=20 and x=20/2=10

y-intercept (initial value)

Set x= 0 in the equation and solve for y: 2(0)-5y=20 and y=20/(-5)=-4

If the equation is in the slope-intercept form y=ax+b

1. y-intercept is b

2. To get the x-intercept ( zero of the function) Set y=0=ax+b, and x=-b/a

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

Definition

A mathematical statement used to evaluate a value. An equation can use any combination of mathematical operations, including addition, subtraction, division, or multiplication. An equation can be already established due to the properties of numbers (2 + 2 = 4), or can be filled solely with variables which can be replaced with numerical values to get a resulting value. For example, the equation to calculate return on sales is: Net income ÷ Sales revenue = Return on Sales. When the values for net income and sales revenue are plugged into the equation, you are able to calculate the value of return on sales.

There are many types of mathematical equations.

1. Linear Equations y= mx + b (standard form of linear equation)

2. Quadratic Equations y= ax^2+bx+c

3. Exponential Equations y= ab^x

4. Cubic Equations y=ax^3+ bx^2+cx+d

5. Quartic Equations y= ax^4+ bx^3+ cx^2+ dx+ e

6. Equation of a circle (x-h)^2+(y-k)^2= r^2

7. Constant equation y= 9 (basically y has to equal a number for it to be a constant equation).

8. Proportional equations y=kx; y= k/x, etc.

Jun 14, 2011 | Computers & Internet

I am not quite sure how the major axis of your hyperbola is directed and i do not know if the lengths you give are measures of the major and minor axes or the measures of the semi-major and semi-minor axes. So I am giving you the equations and the graphs so that you can decide for yourself what is appropriate for your problem.

Major axis parallel to the X-axis

Equation and graph

Center is at x=1 and y=-2, semi-major axis length is a=6, and semi-minor axis length is b=12

Major axis is parallel to the y-axis

Equation

Center is at x=1 and y=-2, semi-major axis length is a=12, and semi-minor axis length is b=6.

I trust you can customize the equations to fit your need.

Major axis parallel to the X-axis

Equation and graph

Center is at x=1 and y=-2, semi-major axis length is a=6, and semi-minor axis length is b=12

Major axis is parallel to the y-axis

Equation

Center is at x=1 and y=-2, semi-major axis length is a=12, and semi-minor axis length is b=6.

I trust you can customize the equations to fit your need.

Mar 24, 2011 | Office Equipment & Supplies

The short story is that this calculator does have a computer algebra system or CAS and thus cannot factor polynomials with arbitrary (unknown) coefficients or known coefficients.

However if the coefficients are given you can ,if you are willing to travel that way, factor approximately a polynomial P(x).

Basically, the idea is that any polynomial P(X) of degree n can be written in the factored form (X-x_1)(X-x_2)...(X-x_n), where x_1, x_2, x_3,...x_n are the roots (real or complex) of the equation P(X)=0.

The procedure ( for a 3rd degree polynomial) is as follows:

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.

However if the coefficients are given you can ,if you are willing to travel that way, factor approximately a polynomial P(x).

Basically, the idea is that any polynomial P(X) of degree n can be written in the factored form (X-x_1)(X-x_2)...(X-x_n), where x_1, x_2, x_3,...x_n are the roots (real or complex) of the equation P(X)=0.

The procedure ( for a 3rd degree polynomial) is as follows:

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.

Sep 11, 2010 | Casio FX-9750GPlus 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

Sep 19, 2010 | Vivendi Excel@ Mathematics Study Skills...

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