math

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math

Postby goffbbb » Tue Mar 16, 2004 7:10 am

A taxi charges $1.20 for the first 1/6 mile and $.40 for each additional 1/6 mile. If the taxi fare was $18.80, how far did the taxi drive?
goffbbb
 

Re: math

Postby Knight » Tue Mar 16, 2004 9:02 pm

goffbbb wrote:A taxi charges $1.20 for the first 1/6 mile and $.40 for each additional 1/6 mile. If the taxi fare was $18.80, how far did the taxi drive?
The best way to solve this is to figure out how many units of distance he drove, then figure out how many miles that is.

Step one: Convert the word problem into mathmatical language.

If we call each unit x, then the first x cost $1.20. There will be x-1 more units at 0.40, until we reach a total of 18.80 That means that:

1.20 + x-1(0.40) = 18.80

Lets simplify the problem now:

We want to solve for x, so we'll move the $1.20 by subtracting it from both sides of the equation.
Code: Select all
 1.20 + x-1(0.40) = 18.80
-1.20               -1.20
-------------------------
        x-1(0.40) = 17.60
Next, we'll deal with the 0.40 by dividing both sides by it.
Code: Select all
x-1(0.40) = 17.60
---------   ------
  (0.40)    (0.40)
Which gives us:
Code: Select all
     17.60
X-1= -----
     0.40
That simplifies to:

x-1 = 44

Finally,
Code: Select all
x-1 = 44
 +1   +1
--------
x   = 45

Now, if each x is 1/6 of a mile, we can obtain the number of miles by multiplying 45 by 1/6:

45 x 1/6

Notice that this simplifies to 45/6 which equals 7.5 miles.
William J. Knight
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Los Alamos National Labs
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Postby Guest » Thu Nov 18, 2004 8:19 am

Given that the matrix A is symmetric if and then find the values of x, y and z (if possible).
Guest
 

Postby Knight » Thu Nov 18, 2004 9:17 pm

Anonymous wrote:Given that the matrix A is symmetric if and then find the values of x, y and z (if possible).
When you ask someone to do your math assignment, you need to give them all of the problem. The previous questions to this one hold the keys to the solution of this one.
William J. Knight
Health Physicist
Los Alamos National Labs
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Knight
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Posts: 309
Joined: Tue Jan 13, 2004 9:43 pm
Location: Los Alamos, NM


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