Hi

Finding the center and radius of the circle: y^2=25-x^2

Standard Form of an Equation of a Circle is 

where Pt(h,k) is the center and r is the radius

 y^2=25-x^2

 x^2 + y^2 = 25  Center (0,0) aand r = 5

 

Hi

Which quadratic equation has solutions of 5 and 7?

The factor theorem states that a polynomial f(x) has a factor (x − k) 

if and only if f(k) = 0

 (x-5)(x-7) = x^2 - 12x + 35

 

Hi

putting equations into the standard form for the conic

completing the squares

1) parabola: x^2-2x+8y+9=0  

   (x-1)^2 = -8y - 8

   (x-1)^2 = -8(y+1)

2)parabola: x^2-2x+8y+17=0

   (x-1)^2 = -8y-16

   (x-1)2 = -8(y+2)

3)parabola: y^2-4y-4x=0

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

  (y-2)^2 = 4(x+1)

4)hyperbola; 12x^2-20y^2-12x+40y-37=0 

     12[(x-.5)^2 - .25]  - 20[(y+1)^2-1] - 37 = 0

     12(x-.5)^2  -3  - 20(y+1)^2 + 20 - 37 = 0

      12(x-.5)^2 - 20(y+1)^2 = 20

        (x-.5)^2/20/12 - (y+1)^2 = 1

           

 

Hi

Nature of solutions:

 y^2= 4/7y+5/3

 y^2 -4/7y - 5/3 = 0

  

 

 

There are NO real solutions: discriminant is a negative number

 

Hi

x^2+6x+9=25

factoring left side

(x+3)^2 = 25

 x+3 = ± 5   |taking the square root of both sides of the EQ

 x = -3 ± 5

  x = -8 and x = 2

 

Hi

(x+2)^2 = 36

 x+2 = ± 6

  x = - 2 ± 6

  x = -8 and x = 4

 

Hi

3x^2+4x = 0

 x(3x+4) = 0

zero-product property



 x = 0

(3x+4) = 0

   x = -4/3

 

Hi

f(x)=(x+2)/3

f(0) = (0+2)/3 = 2/3

f(0) = 2/3

 

Hi

combining like terms

-3 + 8x + 4 + (-10x)

  8x + (-10x) + 4 -3

   -2x + 1   OR 1-2x

 

Hi

Using Cramer's rule to solve the system of equations: 

 x-2y+z= 3

3x+2y+z=-3

2x-3y-3z=-5

 x = -1 ,  y = -1  and z = 2   See below



 
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables


  

  

  

  

  

  

  First let . This is the matrix formed by the coefficients of the given system of equations.

  

  

  Take note that the right hand values of the system are , , and  and they are highlighted here: 

  

  

  

  

  These values are important as they will be used to replace the columns of the matrix A.

  

  

  

  

  Now let's calculate the the determinant of the matrix A to get . To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.

  

  

  

  Notation note:  denotes the determinant of the matrix A.

  

  

  

  ---------------------------------------------------------

  

  

  

  Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix  (since we're replacing the 'x' column so to speak).

  

  

  

  

  

  

  Now compute the determinant of  to get . Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.

  

  

  

  To find the first solution, simply divide the determinant of  by the determinant of  to get: 

  

  

  

  So the first solution is 

  

  

  

  

  ---------------------------------------------------------

  

  

  We'll follow the same basic idea to find the other two solutions. Let's reset by letting  again (this is the coefficient matrix).

  

  

  

  

  Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix  (since we're replacing the 'y' column in a way).

  

  

  

  

  

  

  Now compute the determinant of  to get .

  

  

  

  To find the second solution, divide the determinant of  by the determinant of  to get: 

  

  

  

  So the second solution is 

  

  

  

  

  ---------------------------------------------------------

  

  

  

  

  

  Let's reset again by letting  which is the coefficient matrix.

  

  

  

  Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix  

  

  

  

  

  

  

  Now compute the determinant of  to get .

  

  

  

  To find the third solution, divide the determinant of  by the determinant of  to get: 

  

  

  

  So the third solution is 

  

  

  

  

  ====================================================================================

  

  Final Answer:

  

  

  

  

  So the three solutions are , , and  giving the ordered triple (-1, -1, 2)

  

  

  

  

  Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.

  

  

  

 

Hi

sample of 106 golfers showed that their average score on a particular golf

 course was 87.98 with a standard deviation of 5.39.

A) Find the 90% confidence interval of the mean score for all 106 golfers.

 ME = 1.645[5.39/sqrt(106)] = .8612

 CI: 87.98-.8612 < u < 87.98+.8612

 CI: 86.9988 < u < 88.8412 

B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of 106.

  ME = 1.645[5.39/sqrt(130)] = .7776

 CI: 87.2024 < u < 88.7576

C) Which confidence interval is larger and why?

 confidence interval is larger for the smaller sample size of 106 due to 

the fact the ME is larger



Summary of z values for various confidence intervals

	a	a/2	crtical regions	

90%	0.1	5%	z <-1.645	z >+1.645

95%	0.05	2.50%	z <-1.96	z >+1.96

98%	0.02	1%	z <-2.33	z >+2.33

99%	0.01	0.50%	z<-2.575	z >+2.575

 

Hi, Previously posted

 Mary has 30 hats. She has 5 green, 7 blue, 12 red, and 6 brown. 

 selects two hats at random

P(they are different colors) = 1 - P(both are the same color)

P(both are the same color) = 5/30*4/29 + 7/30*6/29 + 12/30*11/29 + 6/30*5*29

                           = (20 + 42 + 132 + 30)/30*29 =  224/30*29 = .2575

P(they are different colors) = 1 -.2575 = .7425 

 

Hi

 Mary has 30 hats. She has 5 green, 7 blue, 12 red, and 6 brown. 

 selects two hats at random

P(they are different colors) = 1 - P(both are the same color)

P(both are the same color) = 5/30*4/29 + 7/30*6/29 + 12/30*11/29 + 6/30*5*29

                           = (20 + 42 + 132 + 30)/30*29 =  224/30*29 = .2575

P(they are different colors) = 1 -.2575 = .7425 

 

Hi

chart for this quadratic equation 

y=x^2+4x+6

 

Hi

Painter B works twice as fast as A.

Let x and 2x represent time(in hours)it takes solo for B and A

Question states***Per hour the equalizer

 1/x + 1/2x = 1/10  |Multiplying thru by 10x so as all denominators = 1

  10 + 5 = x

     x = 15hr, time solo for B.  time solo for A is 30hr



CHECKING our Answer***

 1/15 + 1/30 = 2/30 + 1/30 = 3/30 = 1/10

 

Hi



solving the quadratic equation

 -3w^2-6w+3=0

factoriing

 -3(w^2+2w-1)= 0

 

Hi, Note the  use of ^ (uppercase 6)

Refer to the Laws governing exponents:

**



 ****





1. 

2. 

3. 

 

Hi

x^2-2x+y^2-15=0   |Only one square to complete

(x-1)^2 -1 + y^2 -15 = 0

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

1) Circle with Center(1,0) and radius of 4

Standard Form of an Equation of a Circle is 

where Pt(h,k) is the center and r is the radius

 

Hi



a. Find the co-ordinates of the vertex and the focus of the parabola 

 x2=4(x+y)

 x^2 - 4x = 4y

 (x-2)^2 - 4 = 4y  OR (x-2)^2 = 4(y+1)

Using the vertex form of a parabola,  where(h,k) is the vertex

 1/4(x-2)^2 - 1 = y   Vertex (2,-1)

 The standard form is , where  the focus is (h,k + p) 

                        (x-2)^2 = 4(y+1)  4p = 4 , p = 1 focus(2,0)

 

Hi

 Standard Form of an Equation of an Ellipse is 

where Pt(h,k) is the center and a and b  are the respective vertices distances from center.

find the foci of the ellipse with the equation:

   | Center(0,0)

 a = 4 and b = 6  f = sqrt(36-16) = sqrt(20) = ±2sqrt(5)

ellipse with foci (0, ±2sqrt(5))

 

Hi

 Standard Form of an Equation of an Ellipse is 

where Pt(h,k) is the center and a and b  are the respective vertices distances from center.

ellipse with foci (0, ± 2) and co-vertices at (± 1, 0).



f =  a^2 = 5

 

Hi

Using the vertex form of a parabola,  where(h,k) is the vertex

y = -4(x+1)^2 +4

 1) parabola opens downward a = -4 < 0  Vertex Pt(-1,4) is a maximum pt

 2) Line of symetry is the x = -1

 3) relative to the the translation from the origin: to the left 1 and up 4

 

Hi

Using the vertex form of a parabola,  where(h,k) is the vertex

y = -3x^2 -4

 1) parabola opens downward a = -3 < 0  Vertex Pt(0,-4) is a maximum pt

 2) Line of symetry is the y-axis

 3) relative to the the translation form the origin: down 4

 

Hi

Using the vertex form of a parabola,  where(h,k) is the vertex

y = (x-4)^2+6

 1) parabola opens upward a = 1 > 0  Vertex Pt(4,6) is a minumum pt

 2) Line of symetry is the x = 4

 3) relative to the the translation form the origin: to the right 4 & up 6

 

Hi

Using the vertex form of a parabola,  where(h,k) is the vertex

f(x)=1/2x^2-6

 1) parabola opens upward a = 1/2 > 0  Vertex Pt(0,-6) is a minumum pt

 2) Line of symetry is the  y -axis

 3) relative to the the translation from the origin: down 6.

 

Hi

Let x represent the time to complete 100% of the work

Question sets up the Proportion***

75%/100% = 85hr/x  Cross Multiplying 

 75x = 85*100

   x = 85*100/75

   x = 113 1/3 hr, the time to complete 100% of the work

 

Hi

Sally can paint a house in 4 hours, and John can paint the same house in 6 hour

Let x represent the time for both of them to paint the house together

Question states*** Per hour the equalizer

  1/4 + 1/6 = 1/x

  3x + 2x = 12   |multiplying thru by 12x so as all denominators=1

     5x = 12

      x = 12/5 or 2.4hr Or 2 hour and 24 min

 

Hi

Note:  The Domain is the allowable values for x

find the domain of the function 

 f(x)= 3/(3x-24)  |CANNOT divide by zero

Denominator = 0 when x = 8

 Domain: {x| x ≠ 8}  Which says x is all Real numbers with the exception of 8.

 

Hi

y varies directly with the square of x and inversely with z:  

 y = kx^2/z

16 = k 2^2/3   | y = 16 when x=2 and z=3

 12 = k

find y when x=5 and z=4

 y = 12*5^2/4

 y = 75

 

Hi

If I am understqanding Your question(s) properly:

2)    OR 

    4(x-1) + 3(x-2) = 2 |Multiplying thru with 12 so as all denominators = 1

        7x = 12

         x = 12/7

3)   |Multiplying thru with 9x so as all denominators = 1

     3*5  - x = 9

        6 = x

4)   | Cross Multiplying 

       4(3x-2) = 3(2x+1)

          6x = 11

           x = 11/6



1)

  Check Your sign: should this be?

 

then

    2 - 6(x-3) = 2(x+4)   |Multiplying thru with (x+4)(x-3):  

    2 - 6x + 18 = 2x+4

        16 = 8x

         x = 2

 

Hi

-21x3+15x2-30x = -3x(7x^2-5x+10)