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Notice that when Car B starts, car A has a 30 mile "lead.' And car B cuts this disttance by 10 miles every hour. So....it takes 3 hours for B to catch A. so....3 hrs. later than 10 AM  = 1 PM.....and that's when B passes A.

11^5=161051?

Awww thank you. :)

I'll start it by continuing off from last time.

$$\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6$$

This is the entire coding correct...

Hopefully it shows...My laptop might not be able to take in all this LaTeX.

\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\
=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\
=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\
=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\
=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\
=(9Cr)(z^{((18-3r)/2)}*(-2)^r)
\\The\:constant\:term\:will\:be\:when\\
(18-3r)/2=0\\18-3r=0\\r=6

omg, its the math god again. i need to hide somewhere!(sarcasm)

only german and aussie cowards run and hide.

the troops from the Empire of Royality are far from ordinary, unlike you guys, we can defeat even gods, if we spawn father david and Mother nature

theyre both with us!

speed=distance/time

distance = speed *time

car 1

distance = 30*t

car 2

distance = 40(t-1)

they will be at the same spot when

30t=40(t-1)      

30t=40t-40

-10t=-40

t=4 hours

t=

Oops, thank you Alan.

I´m to slow ;)

Have a nice day and thanks for the rule!

In mathematics 0/0 is undefined.

Hahaha!

Lol, if I'm the god then I can't describe what Melody is, she taught me this!