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is it 51 degree north of west
A. 20.33 miles B. 43.25 miles C. 28.20 miles D. 22.72 miles E. 37.19 miles
that doesnt seem reasonable......because it said round your answer to two decimal points. oh answers!
Find x and Y component of speed of each ship
ok so ship a would be 28 miles and shib b would be 34 miles i think
Here How I would solve it .. Ship 1 travel at 51 degree North of West Using standard coordinate system X component of speed is -cos(51)*14 Y component of speed is sin(51)*14
This is just ship 1
try this with ship 2
so you will get speed of two ship
Multiply x and Y component of each ship seperately to find position then just use distance formula
ok so ship a is -97.49 ship b is 128.6 ?
X position of Ship 1 at noon(12-10am) -cos(51)*14 mile per hours * 2hour -17.62 Y position of Ship 2 at noon sin(51)*14*2=21.76
X position of Ship 2 -cos(56)*17*2=something Y position of Ship 2 -sin(56)*17*2=something Distance Formula \[\sqrt(ship 1 (x) position - ship 2( x) position)^2+ship 1 (y) position - ship 2( y) position)^2)\]
X position of Ship 2 -cos(56)*17*2=-23.89 Y position of Ship 2 -sin(56)*17*2=17.73
so then I just plug those into the formula?
I got 19.01 for x position of ship 2
hmmm ok you're probably right
can you see the distance formula in full
it is basically square root of (xposition of ship 1- xpos ship 2)^2 + (ypos ship 1-y pos ship 2)^2)
Is it physics or math?
math trig/pre calc
Were you able to get answer
ok I have to go back to the begining and replug in stuff cause i messed something up hold on
I will do the same thing
I got 45.73.....?
The answer is 37.19
I read that problem wrong it is 51 degree west of north ---- not north of west so I have to switch sin to cos vice versa
okay I have no clue what i did then
Ship 1 position X: - sin(51)*14*2 -- A Y: cos(51)*14*2 --B Ship 2 position X: -sin(56)*17*2 -- C Y: -cos(56)*17*2 --D
ok thank you!