Ship A and Ship B are 120 km apart when they pick up a distress call from another boat. Ship B estimates that they are 70 km away from the distress call. They also notice that the angle between the line from ship B to ship A and the line from ship A to the distress call is 28°. What are the two possible distances, to the nearest TENTH of a km, from ship A to the boat?

147.5 km and 64.4 km

Step-by-step explanation:

a=120  km

b=70  km

β=28 degrees ( ∘)

b^2=(a^2)+(c^2)−2ac*cosβ  

70^2 =(120^2 )+(c^2)−2⋅ 120⋅ c⋅ cos(28∘ )  

 (c^2 ) −211.907c+9500=0  

note p, q, and r are replacement variables in the Pythagorean theorem since a, b, and c are already in use

p=1;q=−211.907;r=9500  

D=(q^2 ) −4pr=(211.907^2 )−4⋅1⋅9500=6904.75561996  

D>0  

 =   (−q±     )/2p=(211.91±)/2

 =105.95371114±41.5474295834  

(−147.501140726)(−64.4062815596)=0

=147.501140726  

=64.4062815596  

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.

idk

step-by-step explanation: