recalling that d = rt, distance = rate * time.
we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?

so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.
then Wanda kicks in, rolling like a lightning at 16mph.
let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.
the distance both have travelled is the same "d" miles, reason why they "meet", same distance.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20Hector%26d%2612%26t%2B%5Cfrac%7B3%7D%7B2%7D%5C%5C%5B1em%5D%20Wanda%26d%2616%26t%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D%2812%29%5Cleft%28%20t%2B%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%5C%5C%5B1em%5D%20d%3D%2816%29%28t%29%20%5Cend%7Bcases%7D)

Sounds simple to me but may be misleading, I can't see a trick.
I think all you do is multiply each by 30.
So answer is 45ft X 37.5ft
Hope it's correct. I'd appreciate if you could let me know. Thks..
Answer:
about -1
Step-by-step explanation:
0.9999999999 minus 2
Answer:
C
Step-by-step explanation:
the sum of complementary angles = 90°, hence
x + 5 + x - 9 = 90
2x - 4 = 90 ( add 4 to both sides )
2x = 94 ( divide both sides by 2 )
x = 47
Thus ∠DFG = 47 + 5 = 52° and ∠JKL = 47 - 9 = 38°
Answer:
B.
C.
D.
E.
Step-by-step explanation: