CPhill

Rate of change  = [ final distance -  beginning distance ] / [ final time - beginning time ]

a. Rate of change over the whole trip  =  Miles / Hours  = 350 / 8  = 43.75 mph

b.  Between 4th and 5th hour  =  [300 - 225 ] / [ 5 - 4 ] = 75 / 1  = 75 mph

     Between the  5th and 6th hour  = [300 - 300] / [ 6 - 5]  = 0 / 1  = 0 mph

     Between the   6th and 8th hour = [ 350 - 300 ] / [ 8 - 6 ]  = 50/2  = 25mph

c.  No....if they were....the line would have a constant slope....but it doesn't

d. No...for the same reason given in (c)

7x^2014  * 2x^2012  *  9x^2007   *  8x^2009

___________________________________  =

4x^2013 * 28x^2011 * x^2008  * 6x^2010

7x  * 2x * 9  * 8

________________  =

4 * 28 * x * 6x

1008x^2

_______   =

672x^2

   3

____

   2

It can be shown that the area will be  maximized when each side of the rectangle is equal to the perimeter / 4....in essence the rectangle will be a square

So...each side will  =  44/ 4  =  11 cm

And the maximum area is   11^2  =  121 cm^2

Well...if we wanted to  simplify  inches to feet we have

5 : 85    (simplify by dividing both by  5)

1 (inch) : 17 (feet)     ⇒   Answer 1

If an object that is 85 ft long is represented on a scale drawing as a line 5 in. long, what is the scale used to make the drawing?

85 ft  =   85 * 12  =  1020  inches

So the scale to the real  is :

5 : 1020      (simplify)

1 : 204 

2(x+b) = ax+c    Simplify

2x + 2b   = ax  + c        If this has infinite solutions, then   a  = 2  and 2b  = c

So...note that

2x  + 2b  = 2x + c     will be true for any real x whenever  2b  =  c

If 3x-6y = 9z, which of the following expressions is equivalent to x^2 - 4xy + 4y^2

Dividing the first equation through by 3, we have

x - 2y  = 3z     (1)

Notice that   x^2 - 4xy  + 4y^2   factors as     (x -2y) ( x - 2y)        (2)

So subbing  (1)  into (2)  we get     (3z) (3z)  =  9z^2

Discovering a pattern  is often the key to  solving this type  of problem......

Let's look  at the  first 12  terms....we have

( 2001, 2002, 2003, 2000, 2005, 1998, 2007, 1996, 2009,  1994,  2011, 1992 .....  )

Note that the patern appears to  be that every  4th term decreases the series  by 4 

So we have that

2000  - 0(4)  =  2000  - 0   = 4th term

2000 - 1(4)  = 2000- 4 =  1996  =  8th term

2000 - 2(4)  = 2000 - 8 =  1992  = 12th term

So

2000 - 500(4) =2000 - 2000  = 0 =   2004th term 

This is a right triangle  with AB forming the hypotenuse and AC and BC forming the legs.....the area will =  1/2 the product of the leg lengths...so we have

Area  = (1/2) (AC) ( BC)   =  (1/2)(8)(15)  =  (1/2) (120)  = 60  units^2

So.....letting ....AB  be the base of the triangle and CD  the altitude....then we have that

Area  = (1/2) (17) ( CD)

60  = (1/2) (17) (CD)

120/17 = CD

Now....ΔACB   is similar to  Δ ADC....this means that

AC / CB  = AD/ DC

8/15  = AD / (120/17)

(120/17)(8/15)  = AD

(120/15)(8/17)  = AD

(8)(8) / 17 = AD

64/17  = AD

Since  ADC is also a right triangle  with legs AD and  CD....its area  =  (1/2)(AD)(CD)  = (1/2)(64/17)(120/17)  =

(64 * 60) / 17^2  =

3840/289  units^2  ≈ 13.29 units^2

Thanks, Gavin !!!!....here's another way using  the Law of  Cosines

Here's a pic :

Because AD is an angle bisector, then AC /AB  = CD/BD

So 

4/5 = CD/BD

Then there are 9 parts of BC , then  CD are 4 of them...so CD  = 5 (4/9)  = 20/9

So  BD  = 5 - 20/9 =  45/9 - 20/9  = 25/9

CD^2  = AD^2  + AC^2  - ( AD * 4) cos DAC

(20/9)^2  =  AD^2  + 4^2  - ( AD * 4) cos DAC

(400/81 - 16 - AD^2) / ( - 4AD) = cosDAC     (1)

BD^2 = AD^2 + AB^2  - (AD * 5)cosDAB

(25/9)^2  = AD^2 + 5^2  - (AD * 5)cosDAB

(625/81 - 25 - AD^2) / ( - 5AD) = cos DAB   (2)

Since  angle DAC = angle DAB, their cosines are equal...so...we can equate (1)  and (2)

(400/81 - 16 - AD^2) / ( - 4AD)  = (625/81 - 25 - AD^2) / ( - 5AD) 

Multiply both sides by  -AD

( 400/81 - 16 - AD^2) / 4  = ( 625/81 - 25 - AD^2) / 5

Cross-multiply 

5 ((400/81 - 16 - AD^2) = 4 ( 625/81 - 25 - AD^2)    

Simplify

2000/81 - 80  - 5AD^2  = 2500/81 - 100 - 4AD^2

20 - 500/81 = AD^2

[1620 - 500] / 81  = AD^2

1120/ 81   = AD^2

Note that  1 x 2 X 3 X 4 X 5   =  120

So...any  further multiplications, i.e., 6 x 7 x 8 x.....x 2102 x 2013  will also  end in  "0"