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November 24, 2014

November 24, 2014

Posted by **Jacob** on Tuesday, January 29, 2008 at 10:45pm.

- Algebra1 -
**Anonymous**, Tuesday, January 29, 2008 at 10:47pmDid they leave at the same time?

- Algebra1 -
**Damon**, Tuesday, January 29, 2008 at 10:58pmThey could not have left at the same time. Jacob-you left part of the problem out.

When you get it figured out, the distances are the same

d = rate * time

rate jet1 * time jet 1 = rate jet2 *(time jet 1 - delay)

- Algebra1 -
**Jacob**, Tuesday, January 29, 2008 at 10:58pmno the second one left a half an hour later sorry i forgot to include that

- Algebra1 -
**Damon**, Tuesday, January 29, 2008 at 11:01pmthen

564 *t = 744 * (t - .5)

solve for t, the time for jet 1

then if we want the time for jet 2, subtract 1/2 hour

- Algebra1 -
**Jacob**, Tuesday, January 29, 2008 at 11:09pm564t=744(t-.5)

564t=744t-372

564t-564t=744t-372-564t

0=180t-372

0+372=180t-372+372

372=180t

372/180=180t/180

2.07=t

Is that correct?

- Algebra1 -
- Algebra1 -
**Damon**, Tuesday, January 29, 2008 at 11:17pmYes

- Algebra1 -
**Damon**, Tuesday, January 29, 2008 at 11:22pmBut remember to subtract half hour for second plane

- Algebra1 -
**Jacob**, Tuesday, January 29, 2008 at 11:24pmok thank you for all of your help

- Algebra1 -
- Algebra1 -
**Angelica(:**, Monday, October 10, 2011 at 7:11pm564*2.07=744*2.07-.5

564*2.07=744*1.57

1167.48=1168.08

how come it doesnt equal the same thing?

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