ST is not he midpoint of SN because y=15 and if you plug in you get SN to be 150 but ST is only 64 which is not half of 150
<span>N(t) = 16t ; Distance north of spot at time t for the liner.
W(t) = 14(t-1); Distance west of spot at time t for the tanker.
d(t) = sqrt(N(t)^2 + W(t)^2) ; Distance between both ships at time t.
Let's create a function to express the distance north of the spot that the luxury liner is at time t. We will use the value t as representing "the number of hours since 2 p.m." Since the liner was there at exactly 2 p.m. and is traveling 16 kph, the function is
N(t) = 16t
Now let's create the same function for how far west the tanker is from the spot. Since the tanker was there at 3 p.m. (t = 1 by the definition above), the function is slightly more complicated, and is
W(t) = 14(t-1)
The distance between the 2 ships is easy. Just use the pythagorean theorem. So
d(t) = sqrt(N(t)^2 + W(t)^2)
If you want the function for d() to be expanded, just substitute the other functions, so
d(t) = sqrt((16t)^2 + (14(t-1))^2)
d(t) = sqrt(256t^2 + (14t-14)^2)
d(t) = sqrt(256t^2 + (196t^2 - 392t + 196) )
d(t) = sqrt(452t^2 - 392t + 196)</span>
Check the picture below.
![~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{-9})\qquad B(\stackrel{x_2}{8}~,~\stackrel{y_2}{0}) ~\hfill AB=\sqrt{[ 8- 1]^2 + [ 0- (-9)]^2} \\\\\\ AB=\sqrt{7^2+(0+9)^2}\implies AB=\sqrt{7^2+9^2}\implies \boxed{AB=\sqrt{130}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20distance%20between%202%20points%7D%7D%7Bd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%7D~%5Chfill~%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-9%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B0%7D%29%20~%5Chfill%20AB%3D%5Csqrt%7B%5B%208-%201%5D%5E2%20%2B%20%5B%200-%20%28-9%29%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AB%3D%5Csqrt%7B7%5E2%2B%280%2B9%29%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B7%5E2%2B9%5E2%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B130%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![B(\stackrel{x_1}{8}~,~\stackrel{y_1}{0})\qquad C(\stackrel{x_2}{9}~,~\stackrel{y_2}{-8}) ~\hfill BC=\sqrt{[ 9- 8]^2 + [ -8- 0]^2} \\\\\\ BC=\sqrt{1^2+(-8)^2}\implies \boxed{BC=\sqrt{65}}](https://tex.z-dn.net/?f=B%28%5Cstackrel%7Bx_1%7D%7B8%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_2%7D%7B9%7D~%2C~%5Cstackrel%7By_2%7D%7B-8%7D%29%20~%5Chfill%20BC%3D%5Csqrt%7B%5B%209-%208%5D%5E2%20%2B%20%5B%20-8-%200%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20BC%3D%5Csqrt%7B1%5E2%2B%28-8%29%5E2%7D%5Cimplies%20%5Cboxed%7BBC%3D%5Csqrt%7B65%7D%7D)
now, we could check for the CA distance, however, we already know that AB ≠ BC, so there's no need.
Assuming you are focusing on the square only, then the answer is 9 inches.
You take the square root of 81 to get 9. Notice how 9^2 = 9*9 = 81. So if you're not familiar with square roots, then think backwards in a sense to get the answer.
Answer:
−45−1+3−63−6
−46−1+3−63−6
−46+2−63−6
Solution:
−46−63−6+2
Step-by-step explanation:
I hope that this is what you are looking for
Have a nice day/night
