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Jessica Malone

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I think tangent lines are drawn on your graph at these points......the SLOPES of the lines will be the velocity at that point.

At 2 seconds you have a line with slope   rise/run     50/2 = 25 m/s    approx

at 4 seconds the tangent line is 95/2 = 47 m/s    approx.

    going faster after 4 seconds due to acceleration.....

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What's the question.....???

60 ( 1 - .11)^3   ≈  $42.30

Exponential (decay)

Oh right thanx!

The instantaneous velocity at any point will simply  be the slope of the line through that point

For instance....look at the line through the point that represents the position after 2 sec

The following approximate points are on this line  (1,0)  and ≈ ( 3.5, 60)

So....the approximate slope  is  [ 60 - 0 ] / [ [3.5 - 1 ] =  60 /2.5   ≈  24m / s

And look at the line through the point that represents the position after 4  sec

The following approximate points are on this line  (2,0)  and ≈ (4.5, 120)

The approximate slope is   [120 - 0 ] / [ 4.5 - 0  ] = [120/4.5]  ≈  26.67 m/s

This makes sense.....the slope at 4 sec  is steeper than at 2 sec....so....we would expect the instantaneous velocity at 4 sec should be  > than at 2 sec

Thank you verry much Travisio for your amazing response

One more thing about matrix multiplication    .....the resulting matrix elements will tell you what row, column of each matrix to multiply to get a partcicular element

For  example.....we are multiplying  a 2 x 2  matrix by a 2 x 2 matix

The resulting matrix  will be a 2 x 2  with the following elements

 [   a ₁₁    a ₁₂

     a ₂₁   a  ₂₂  ]

This tells you that  the   element   a _mn   is obtained by multiplying the mth row of the first matrix by the nth column of the second

So....for example.....element a ₁₁   is obtained by multiplying the first row of the first matrix by the first column of the second.....etc.

A  =  3 x 2

B  =  2 x 2

C =   1 x 3

D  =   2 x 1

Here's a rule        we can  multiply  martices  only   with the same  "inner" dimensions

And the resulting matrix wll be  the "outer" dimensions

For  example

Matrx 1    =    a  x  b

Matrix 2  =     b  x  c 

So

     M1     *     M2       =

[ a x b ]   *  [ b * c ]

We can multiply these because the  "inner dimensions"  [ the blue b's]  are the same

And the resulting matrix will have the dimensions of  a x c    [ the "outer" dimensions ]

So

BD  =   [2 x 2 ] * [ 2 x 1 ]  =  [ 2 x 1 ]

CA  = [ 1 x  3 ] * [ 3 x 2 ]    =  [ 1 x 2 ]

DC  = [  2 x1 ] *  [ 1 x 3 ]   =  [  2 x 3  ]

ST   =

 3   4        - 2  0      =       [ -6 - 16      0 + 20       =    [ - 22    20

-1 - 2       -  4  5                 2 + 8      0 +  10 ]                10   -10  ] 

TS  =

-2   0       3   4       =         [ -6 + 0      -8 + 0           =    [  -6      -8

-4  -5      -1  -2                    -12 + 5    -16 + 10 ]              -7     -6 ]

They are not the same   [as we should expect, in general ]

BTW, Jenny.....here is an excellent site to check your work with these  : 

http://matrix.reshish.com/multiplication.php

Thanks , CU and Melody..... (.I do not see how to find the answer, either.....!!! )

What you have written is :

8 + 5 - (2/6)* 4c  =  8       simiplify

13 - (2/6) * 4c  = 8     subtract 13 from both sides

-(2/6)* 4c  =  -5      divide both sides by 4

-(2/6) c  =  -5/4     multiply both sides by -(6/2)

c  =  (-5/4)(-6/2)   =   30/8   =  15/4   =  3,75

c=3.75

log ( 118652.56484) =

x > 5.074277130533296