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Posted on Jan 02, 2017

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Try casting the equation of the circle into the form (x-h)^2+(y-k)^2=R^2. Here (h,k) would be the center and R the radius.

To do that complete the two squares.

x^2-2x=x^2-2x+1-1=(x^2-2x+1)-1=(x-1)^2 -1.

Do the same for the y terms.

y^2-2y=(y-1)^2 -1.

Substitute these two expressions in your original equation.

(x-1)^2+(y-1)^2 -1-1=14 or (x-1)^2+(y-1)^2=14+2=16

The radius is square root of 16 or 4, and the center is C(1,1).

To do that complete the two squares.

x^2-2x=x^2-2x+1-1=(x^2-2x+1)-1=(x-1)^2 -1.

Do the same for the y terms.

y^2-2y=(y-1)^2 -1.

Substitute these two expressions in your original equation.

(x-1)^2+(y-1)^2 -1-1=14 or (x-1)^2+(y-1)^2=14+2=16

The radius is square root of 16 or 4, and the center is C(1,1).

Sep 20, 2011 | SoftMath Algebrator - Algebra Homework...

It's going to be a line on which every point represents values of x and y where 2*x is 6 more than 8*y.

Aug 18, 2011 | Computers & Internet

- This equation is in standard form. For more information, watch this:

Jul 07, 2011 | Computers & Internet

First step is to find inequality in the form y<f(x), in this case **y<5-0.5x**. Then graphing it. See captured image

Feb 18, 2011 | Texas Instruments TI-84 Plus Silver...

Use the rule for differentiating products of functions: ()' signifies derivative

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form

sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]

then calculated the derivative of

29/2*[cos(X)-cos(3X)]

which is

29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent

29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

(29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* (sin(2X))'*sin(X) +29*sin(2X)*(sin(X))'

But

- (29)'=0 derivative of a constant is zero
- (sin(2X))'=cos(2X)*(2X)'=2*cos(2X)
- (sin(X))'=cos(X)

(29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)

You could also have cast your formula in the form

sin(2X)*sin(X)= 1/2[ cos(2X-X)-cos(2X+X)]=1/2[cos(X)-cos(3X)]

then calculated the derivative of

29/2*[cos(X)-cos(3X)]

which is

29/2*[-si(X) +3*sin(3X)]

The challenge for you is to prove that the two forms are equivalent

29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X)=29/2*[-si(X) +3*sin(3X)]

Jun 21, 2010 | Vivendi Excel@ Mathematics Study Skills...

Hello,

You are talking of the distributivity of multiplication with respect to the addition or subtraction. The only term where you have multiplication and addition/subtraction is the one with parentheses because

4(5-6y) = 4*(5-6y). * means multiply by

Distributing the product consists in removing CORRECTLY the parentheses.

I consider the term alone

4*(5-6y) = 4*5 +4*(-6y)= 20 -4*6*y = 20 -24*y

You put this result in a package and replace it where it was it the original expression.

So

8*y -(package) +20= Replace package

8*y -(20-24*y)+20 = 8*y -20 -(-24y) + 20 Remember -*(-) = +

8*y -20 +24*y +20= Remember -20+20 =0

8*y -20 +20 +24*y= Addition is commutative

8*y +24*y = 32*y

Let y be an orange 8 oranges + 24 oranges =32 oranges.Coclusion

**8y-4(5-8y)+20 = 32y**

You are talking of the distributivity of multiplication with respect to the addition or subtraction. The only term where you have multiplication and addition/subtraction is the one with parentheses because

4(5-6y) = 4*(5-6y). * means multiply by

Distributing the product consists in removing CORRECTLY the parentheses.

I consider the term alone

4*(5-6y) = 4*5 +4*(-6y)= 20 -4*6*y = 20 -24*y

You put this result in a package and replace it where it was it the original expression.

So

8*y -(package) +20= Replace package

8*y -(20-24*y)+20 = 8*y -20 -(-24y) + 20 Remember -*(-) = +

8*y -20 +24*y +20= Remember -20+20 =0

8*y -20 +20 +24*y= Addition is commutative

8*y +24*y = 32*y

Let y be an orange 8 oranges + 24 oranges =32 oranges.Coclusion

Sep 10, 2009 | The Learning Company Achieve! Math &...

4x+8y=20 || /4

x+2y=5

x=5-2y

-4x+2y=-30 || /2

-2x+y=-15

-2(5-2y)+y=-15

-10+4y+y=-15

5y=-5

y=-1

x=5-2y || y=-1

x=5-2(-1)

x=5+2

x=7

x+2y=5

x=5-2y

-4x+2y=-30 || /2

-2x+y=-15

-2(5-2y)+y=-15

-10+4y+y=-15

5y=-5

y=-1

x=5-2y || y=-1

x=5-2(-1)

x=5+2

x=7

Mar 10, 2009 | Bagatrix Algebra Solved! 2005 (105101) for...

use the simultaneous equation solver software available on ur ti-89 and enter the equation in the form of augmented matrix and you will get ur answer.

Dec 03, 2008 | Texas Instruments TI-89 Calculator

V (2,0) and Focus (2,2)

since the focus is 2 units above the vertex, the parabola opens upward

vertex (h,k)

a is the focal length (distance between the vertex and the focus)

a = 2

(2 units above the vertex)

(x-h)^2 = 4a (y-k)

(x-2)^2 = 4(2) (y-0)

x^2 - 4x + 4 = 8 (y)

x^2 - 4x + 4 = 8y

x^2 - 4x - 8y + 4 = 0

**Answer: "(x-2)^2 = 8 (y)" or in expanded form "x^2 - 4x - 8y +4"

since the focus is 2 units above the vertex, the parabola opens upward

vertex (h,k)

a is the focal length (distance between the vertex and the focus)

a = 2

(2 units above the vertex)

(x-h)^2 = 4a (y-k)

(x-2)^2 = 4(2) (y-0)

x^2 - 4x + 4 = 8 (y)

x^2 - 4x + 4 = 8y

x^2 - 4x - 8y + 4 = 0

**Answer: "(x-2)^2 = 8 (y)" or in expanded form "x^2 - 4x - 8y +4"

Nov 18, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

Consider the following system of **3 equations in 3 unknowns**:

*x + y = *2

**2***x + *3*y + z = *4

*x + *2*y + *2*z = *6Our goal is to transform this system into an equivalent system from which it is easy to find the solutions. We now do this step by step.
* x + y = *2

* y + z = *0

* y + *2*z = *4
*z = *4, *y = *-4, and *x =* 2*-*(*-4*)* = *6Equivalently, we say that the unique solution to this system is **(***x, y, z*) = (6, -4, 4).

- Subtract 2*(Row1) from Row2 and place the result in the second row; subtract Row1 from Row2 and place in the third row. Leave Row1 as is.

- Subtract Row2 from Row3, and place the result in row3. Leave Row1 and Row2 as they are.

Sep 17, 2008 | Belkin (F5D7230-4) Router (587009)

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