CPhill

We have that

200 = 250(.5)^(12/h)   where h is the half-life in hours

Divide both sides by 250

200/250  = (.5)^(12/h)

.8  = (.5)^(t/h)      take the log of both sides

log (.8)  = log(.5)^(12/h)   

And by a log property we can write

log (.8) =  (12/h) log (.5)

Rearrange to isolate h

h  = 12log(.5) / log(.8)  ≈  37.28 hrs   

Let's work backwards

y  = x^2/4  +  x + 4     subtract 4 from both sides

y - 4  =  x^2/4 + x      factor out the 1/4 on the right side

y - 4   = (1/4)(x^2 + 4x)    complete the square on x within the parentheses

y - 4  = (1/4)(x^2 + 4x  + 4  - 4)   factor the first three terms inside the parentheses

y - 4  = (1/4) [ (x + 2)^2 - 4 ] distibute the 1/4 across the brackets

y - 4  = (1/4)(x + 2)^2  -1     add 1  both sides

(y - 3)  = (1/4) (x + 2)^2       

In the form (y - k) = a(x - h)^2....the vertex is at  ( -2, 3)

Multiply both sides by 4

4(y - 5) = (x + 2)^2

In the form 

4p (y - k)  = (x - h)^2

p  = the focal distance from the vertex

So...in our equation    4  = 4p   ⇒  p  = 1

Since the parabola turns upward....the focus is given by (h, k + p)  = (-2,  3 + 1) =

(-2, 4)

The directrix is given by   y  = k - p   =  3 - 1  =  2

Look at the graph here : https://www.desmos.com/calculator/wixfeughs1

Both functions are exactly the same

The vertex  is  (-2, 3)

The focus is (-2,4)

The directrix  is  y  = 2

37

x^2 - 16 x  + 64

It will touch the x axis  when y  = 0

So

x^2  - 16x  + 64  = 0     factor

(x - 8)^2  =  0      take the square root of both sides

x - 8  = 0

x  =  8

38

8x*2 + 10x  - 2   = 0

x  =   ( -10 ±√ [ 10^2  - 4(8)(-2) ]  ) /  [ 2*8 ]

x  =  ( -10±√ [ 100 + 64])  / (16)

x  = (-10 ±√164) / 16

x  =  (-10 + √164 ) / 16

x = 0.18

f(x) =  (x^2+4x-12) / ( x-2 )

This will be undefined (have a hole )  for the x value  that gives the form  0/0

Factor the numerator

[(x + 6) (x - 2) ] / (x - 2)

Nothe that when  = 2, we will have the form  0/0

So.....to find the y coordinate of the hole,  cancel the ( x - 2) factors and sub  2  into  x + 6

So.....the coordinate of the  "hole"  is    (2 , 2 + 6)  =  (2,8)

Transformation Rule  =  ( x + 6, y - 8)

Scale factor  =   Radis of circle 2  / Radius of circle 1   =  9 / 6   =   3/2  =  1.5

3

The Mean  is  111.16

The SD  is  32.8

The minimum value is 69

The max value is 158

So  

Two SDs  above the mean  is   111.16 + 2(32.8)   = 176.76

Two SDs  below the mean is  111.16  - 2(32.8)  = 45.56

So  all the values fall within 2SDs  from the mean

4.

I think this is correct NSS

The mean  is 28.3

The SD is  9.93

So....two SDs above the mean  is 28.3 * 2(9.3)  = 46.9

And two SDs below the mean is  28.3  - 2(9.3)  =  9.7

So...

16  is the minimum value   and 40  is the max value

So...all   values fall within  2 SDs  from the mean

Sincs RS  =  5.....then PQ  also equals 5 since RS  is parallel to PQ

So using the perimeter.... PS  + QR  =  24 - 5 -  5  =   14

And since PS = QR  then 

2OR  = 14      divide both sides by 2

QR  = 7

And the area of  PQRS  = ( Base * Height)   = (QR * Height)

So

28  =  7 * Height       divide both sides by 7

4   =  Height    =  ST

And by the Pythagorean Theorem

RT  = sqrt  [ RS^2  - ST^2]  =   sqrt [ 5^2  - 4^2]  = sqrt [ 25 - 16 ] = sqrt (9)  =   3

So....the base  of QOST  = QT =  QR + RT   =  7 + 3   =  10

So  the area of QOST  =  

QT  * ST  =

10 * 4   =

40 units^2

y  =  3 + 4x^2  -  x^3

The derivative of this is the slope at any point, a

So

y'  =  0 + 8x  - 3x^2        simplify

y'  = -3x^2 + 8x

so...the slpoe at x  = a   is

-3(a)^2   + 8a

2

Mean   =  [  53 + 51  + 48 + 49 + 37 ]  / 5    = 47.6

Variance  =  [29.2 + 11.6 + 0.2 + 2 + 112.4 ]  / 5  = 31.1

SD =  sqrt (31.1)   =  5.6

First answer