Question

Can anyone help me with solving real world problems for Algebra II ? Please & thank you .

Answer 1

is that your question?

Answer 2

lol, no.. this is one of my questions :
A plane travels from Orlando to Denver and back again. On the five-hour trip from Orlando to Denver, the plane has a tailwind of 40 miles per hour. On the return trip from Denver to Orlando, the plane faces a headwind of 40 miles per hour. This trip takes six hours. What is the speed of the airplane in still air?

Answer 3

well d=constant in both the cases from orlando to denever and back again so just equate two distances you'll get speed of airplane

Answer 4

Hartnn: I believe so ..
& Ghazi: ... how would I do that?

Answer 5

well i am getting speed of airplane = 40 miles/hour is it correct lol..

Answer 6

@amistre64 help me to figure this out

Answer 7

Now since distance covered is equal,we will equate speed* time for both
Let S be the speed of airplane in still air,and W=40 be the speed of the wind
for 1st journey,total speed S1= S+W ,total time T1=5
for 2nd journey,total speed S2=S-W,total time T2=6
As speed = distance/time
and since distances are equal, S1*T1=S2*T2
so (S+40)*5=(S-40)*6
can u solve for S now?? @gabbytheflower

Answer 8

assuming the same route is taken to and fro; distance remains constant

distance = (unitspeed + 40) * 5
distance = (unitspeed - 40) * 6

Answer 9

80*5 = 0*6 ? nah

Answer 10

what's this?

Answer 11

"well i am getting speed of airplane = 40 miles/hour"

(40+40)*5
(40-40)*6

just saying :)

Answer 12

at least theres no yeild signs up there ....

Answer 13

ohkay :)

Answer 14

:))