geno3141

You have a point and a slope; use the point-slope form:  y - y1  =  m(x - x1)

m = 1 1/2  =  3/2               x1 = -4             y1 = -4

  y - y1  =  m(x - x1)     --->     y - -4  =  (3/2)(x - -4)

                                              y + 4  =  (3/2)(x + 4)

multiply both sides by 2:       2(y + 4)  =  3(x + 4) 

simplify:                                2y + 8  =  3x + 12

subtract 8 from both sides:            2y  =  3x + 4

subtract 3x from both sides:     -3x + 2y  =  4

multiply by -1:                        3x - 2y  =  -4

Because ac, ab, and bc are unlike terms, they can't be combined.

(50 - 2)!  =  48!

50! / [ 2! · (50 - 2)! ]    =    50! / [2! · 48!]

Since 50!  =  50 x 49 x 48 x 47 x 46 x 45 x ... x 3 x 2 x 1   <---  in the numerator

and   48!  =                 48 x 47 x 46 x 45 x ... x 3 x 2 x 1   <--- in the denominator

50! / 48!  reduces to 50 x 49

Also:  2!  =  2 x 1  = 2

After cancelling out, we have    (50 x 49) / (2)  =  25 x 49  =  1225

Your answer for the second an fourth problems are correct (the one that have a "<" sign).

The other problems are incorrect because the calculator that you are using (or the way that you are using it) gives answers for "<"; to get answers for ">", subtract that answer from 1.0000.

For example, your last answer, 0.0384, should be subtracted from 1.0000 to get 0.9616. This is also true for problems 1 and 3.

Find both the first derivative and the second derivative of that function.

If the first derivative is zero for some value, then there is a possible minimum or maximum at that point.

If the function is f(x) and c is the value:

If f'(c) = 0 and f''(c) < 0, then there is a local maximum at x = c.

If f'(c) = 0 and f''(c) > 0, then there is a local minimum at x = c.

CPhill:

If you had the string go from the upper-left corner to the lower-right corner and then twisted the sheet into a cylinder, wouldn't the string just wrap around once?

To wrap around four times, wouldn't you have to go from the upper-left corner to one-fourth of the way down the right side and then repeat that three more times?

For one-fourth of the way, width = 16 and height = 12:

√( 16² + 12²)  = 20

For four times around:  20 x 4  = 80 cm.

f(x)  =  [2 - 3x] / [5 - 2x]

f(a)  =  [2 - 3a] / [5 - 2a]  =  3

Multiply both sides by 5 - 2a:

             [2 - 3a]  =  3[5 - 2a] 

                2 - 3a  =  15 - 6a

Add 6a to both sides:

               2 + 3a  =  15

                     3a  =  13

                       a  =  13/3

You can use the distance formula:  d  =  √[ (x2 - x1)² + (y2 - y1)² ]

(x1, y1) = (0, 5)     --->     x1 = 0   and   y1 = 5

(x2, y2) = (3, 10.5)   --->   x2 = 3   and   y2 = 10.5

d  =  √[ (3 - 0)² + (10.5 - 5)² ]   =   √[ (3)² + (5.5)² ]   =  √[ 9 + 30.25 ]    =   √[ 39.25 ] 

The square of a number is that number times itself.

6²  6 x 6  =  36