SOLUTION: Dear yenigelabert@yahoo.com, your question for free tutors has been submitted. Note, if you need an IMMEDIATE answer and can pay a modest amount, you can go to http://

Question 459816:
Dear yenigelabert@yahoo.com, your question for free tutors has been submitted.
Note, if you need an IMMEDIATE answer and can pay a modest amount, you
can go to
http://www.algebra.com/cgi-bin/secretoffer.mpl?name=liveperson
They have great tutors and cheap rates. Note that algebra.com is completely free.

--------------- QUESTION FOLLOWS --------------------------
if 4 -3a=7-2(2a+5) evaluate a^2+7a
Please I NEED HELP
(-2x^3+x-1)-(-x^2+x-3)
(2x^2+5x-3)-(3x^3+2x-5)
a ship traveling east at 25mph is 10mi from a harbor when another ship leaves the harbor traveling east at 35mph. How long does it take the second ship to catch up.?

Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
if 4 -3a=7-2(2a+5) evaluate a^2+7a
4-3a=7-4a-10
-3a+4a=7-10-4
a=-7
a^2+7a= 49-49
=0
.....
(-2x^3+x-1)-(-x^2+x-3)
-2x^3+x-1 +x^2-x+3
-2x^3+x^2+3

(2x^2+5x-3)-(3x^3+2x-5)
-3x^3+2x^2+3x+5

a ship traveling east at 25mph is 10mi from a harbor when another ship leaves the harbor traveling east at 35mph. How long does it take the second ship to catch up.?

catch up distance = 10 miles
catchup speed = 35 -25 = 10 mph
catchup time = catchup distance / catch up speed
10/10
=1 hour to catch up

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