geno3141

This can be done several ways; one way is purely by algebra:

The equation of a circle is x² + y² + Ax + By + C = 0.

For the point:  (-2,2)   --->    x = -2  and  y = 2

     substituting:   (-2)² + (2)² + A(-2) + B(2) + C  =  0

          --->               4 + 4 - 2A + 2B + C  =  0

          --->                         - 2A + 2B + C  =  -8        <---  Equation #1

For the point:  (2,-2)   --->    x = 2  and  y = -2

     substituting:   (2)² + (-2)² + A(2) + B(-2) + C  =  0

          --->               4 + 4 + 2A - 2B + C  =  0

          --->                           2A - 2B + C  =  -8     <---   Equation #2

Putting these two equations together:     -2A + 2B + C  =  -8

                                                                 2A - 2B + C  =  -8

Adding down:                                                          2C  =  -16     --->   C  =  -8

Putting this back into the two equations gives us:

     Equation #1:  -2A + 2B  =  0  --->   -A + B = 0

     Equation #1:   2A - 2B  =  0   --->    A - B  =  0

For the point:  (6,2)   --->    x = 6  and  y = 2     and  C = -8

     substituting:   (6)² + (2)² + A(6) + B(2) + C  =  0

          --->              36 + 4 + 6A + 2B + -8  =  0

          --->                                  6A + 2B  =  -32      --->  3A + B  =  -16

Putting this with the equation #2:                                       A - B  =  0

Adding down the columns:                                              4A  =  -16   --->   A  =  -4

Substituting this back into  A - B  =  0     gives  -4 - B  =  0     --->     B  =  -4

Placing  A = -4,  B = -4,  and  C = -8  into the equation  x² + y² + Ax + By + C = 0

gives  x² + y² - 4x - 4y - 8 = 0.

Are you solving these problems by table or by calculator?

If you use a TE-84+, press: (2nd) (DISTR) 2:normalcdf

Then enter:  normalcdf(88,9000,73,8)

The entries are lower bound, upper bound, mean, standard deviation.

I used 9000 for an upper bound, you can use any ridiciously large number.

If you need to solve this in a different manner, repost.

If you divide the total cost by the cost per kilogram, you will get the number of kilograms.

$1.20 / $0.75  =  1.6 kilograms.

Are you using a table or a calculator?

If you are using a TI-84+ calculator, press (2nd) (DISTR) and 2:normalcdf(

To find P(z<0.52), you can used normalcdf(-100,0.52) to get . 0.698.

To find P(z>-0.63), you can use normalcdf(-0.63,100) to get 0.736.

To find P(-0.31<z<1.345), you can use normalcdf(-0.31,1.345) to get 0.529.

(The answers are rounded.)

You needn't use -100 for the left-hand limit or 100 for the right-hand limit, but you should use numbers <-6 or >+6 for the limits.

Use the formula:  A  =  Pe^rt

where     A = final amount = 20 912     P = initial amount = 2 969

               r = rate (as a decimal)           t  =  time  = 25

Substituting:     20 912  =  2 969e^25r

Dividing both sides by 2 969:     7.043434  =  e^25r

Take the ln of both sides:         ln(7.0434)  =  ln(e^25r)

                                                  ln(7.0434)  =25r

Divide by 25:     r  =  ln(7.0434) / 25     --->     r  =  0.078     --->     r  =  7.8%  (approx)

Sequence:     n ---> 2n + 20.

To find the value of the fifth term,  n = 5

                     5 --->  2(5) + 20     --->     10 + 20     --->     30

x - 7y  =  2          and         x²  =  34y² + 7xy  - 16

Take the linear equation (x - 7y  =  2) and solve it for either  x  or  y.

It seems easier to solve it for x:  x  =  7y + 2

Replace this into the other equation:

  x²  =  34y² + 7xy  - 16     --->       (7y + 2)²  =  34y² + 7(7y + 2)y  - 16

Simplify:     --->     49y² + 28y + 4  =  34y² + 49y² + 14y - 16

Simplify:     --->     0  =  34y² - 14y - 20

Divide both sides by 2:     0  =  17y² - 7y - 10

Factor:                             0  =  (17y + 10)(y - 1)

Solve:                                       y  =  -10/17     or     y  =  1

Substituting these values back into the equation:  x - 7y  =  2

     x - 7(-10/17)  =  2     --->     x  =  -36/17      --->     Answer:   (-36/17, -10/17)

     x - 7(1)  =  2             --->     x  =  9              --->     Answer:    (9, 1)

I believe that when graphed   x²  =  34y² + 7xy  - 16  is a hyperbola  and  x - 7y  =  2  is a straight line, so that their solution, their intersection, will either by no points, one point, or two points; in this case:  two points.

Perhaps someone can post the graph ...

If y = 60, then f(x) = 60.

Since f(x)  =  2x² - 2x - 24,  60  =  2x²​ - 2x - 24

Subtracting 60 from both sides:  0  =  2x²​ - 2x - 84

Dividing both sides by 2:             0  =  x² - x - 42

Factoring:                                   0  =  (x - 7)(x + 6)

Soving for x:                               x = 7  or  x = -6

To solve when the unknown is an exponent, find the log (either base ten or base e) of both sides:

y = 3^x

log(y)  =  log(3^x)

log(y)  =  xlog(3)                          <since exponents become multipliers>

log(y) / log(3)  =  x                       <divide both sides by log(3)>

You can use either log or ln; if you used ln, your answer is:  ln(y) / ln(3)  =  x

Going from one corner of a rectangular box to the diagonally opposite corner:

(distance through box)²  =  (length of box)² + (width of box)² + (height of box)²

In a three-dimensional graph:

d = distance                     x, y, z  = axes

d²  =  x² + y² + z²