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Thanks! 

Melody, will you please help me now that i edited it?

Where is 'd' a minumum?   (this is basically pythagorean theorom)

d = sqrt( 22500t^2 + 225 +160000t^2 - 12000t)      This is a quadratic equation.....

(1)  d^2 = ( 182500t^2 -12000t + 225)                             find the minumum....take the derivative and look for slope = 0

            365000t - 12000 = 0

                   t = .032876 hr

Substitute this value in to the 'd' equation

d = sqrt   (  150 (.032876))^2    +  (15 - 400(.032876))^2  )  = 5.2668 miles

Or take equation (1) and graph it to find the minimum point of the parabola......

Hi Darkside :)

Here you go Rom.

Some gourmet popsicles just for  you      

Angle FED and CEB are vertical angles and therefore congruent. 
Angle EDF and ECB are both right angles and there congruent.

Using the AA Similiarity Postulate, we can say that triangles FED and BEC are similar.
 

Now, we can use the idea that corresponding sides of similar triangles are proportional to find DF. 
First, we know that the sides of the square have lengths of  7 as DE = 2 and EC = 5, so DC = 7.
So, side BC = 7 as well.
Set up a proportion:
\(\frac{DE}{CE}=\frac{DF}{CB}\\ \frac{2}{5}=\frac{DF}{7}\\ \frac{14}{5}=DF\)
DF = 14/5 or 2.8

Thank you once again CPhill

You are right, thank you CPhill 

For latex, use the latex box, but don't include the dollar signs!

Where are you getting this accurate diagram, when taking a screenshot? Aops?

125÷27=4.629 repeating.

hey... where are my popsicles?

125 / 27 = 4.63

Can't you put that into a calculator?

What do you mean? Its LaTeX. Doesn't it work on this? I've seen people use LaTeX on this website?

Am I supposed to take a picture of it in LaTeX?

 Δ EDC   ~  Δ  EBA

So

EC  / DC  = EA / BA

EC / 8  = 5 / 4       multiply both sides by 8

EC  =  8 *  ( 5 /4)  

EC =  40 / 4  

EC  = 10

hi what school do you go to?

You can convert all this to base 10 and solve it but to me that goes against the grain of the problem.

All the following is assumed to be in base 6

AB+C = C0

B<6, C<6, so B+C=6, This produces a carry of 1 to the left so

A+1 = C

AB+BA=CC so B+A = C mod 6

B+A=A+1 mod 6

B=1

C=6-B = 5

A=C-1 = 4

ABC = 415₆

We have that  ΔADE  ~ ΔABC

AB  =  AE + BE   =  11+ 10  = 21

So

AD / AE  = AB / AC

AD / 11  = 21 / 35       multiply both sides by 11

AD   =  11 *  (21 / 35 )

AD  =  11 * ( 3 / 5)

AD  = 33 / 5

Yes, thanks guys :))

What rubbish, this is not your answer.

Shreyas1,

I think that you have copied this and your other answers off another site.

I have seen no evidence that you even know how to do your own Latex.

I would not be surprised if you do not even know what LaTex is.

Thanks!! But I am still confuse what “inverse” really is and does in this problem...

Yes it is, thank you so much

OQ / (OM + OQ)  =  OP ( OP + ON)

6/(9+6) = OP / (25)

6 / 15  =  OP /25

OP = 25  * 6/15

OP = 10

Yep it is 

False for no 1

True for no 2

True for no 3

False for no 4

Ask me later if you want me to show work

Thank you both of you(CPhill and ElectricPavlov)