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schepotkina [342]
3 years ago
14

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 awa

y 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?
Mathematics
1 answer:
oksian1 [2.3K]3 years ago
5 0

Answer:

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  

 

c_{1,2}  =   (−q±  \sqrt{D}   )/2p=(211.91±\sqrt{6904.76})/2

​c_{1,2}  =105.95371114±41.5474295834  

(c_{1}−147.501140726)(c_{2}−64.4062815596)=0

c_{1}=147.501140726  

c_{2}=64.4062815596  

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Please help!? 5 Stars! :0 :(
zvonat [6]

OK.  These problems are easy if you know the quadratic formula,
and they're impossible if you don't.

Here's the quadratic formula:

When the equation is in the form of      Ax² + Bx + C = 0

then                    x =  [ -B plus or minus √(B²-4AC) ] / 2A

I'm sure that formula is in your text or your study notes,
right before these questions.  You should cut it out or
copy it, and tape it inside the cover of your notebook.
Then, you'll always have it when you need it, until
you have it memorized and can rattle it off.

The first question says        3x²  +  5x  +  2  = 0

Is this in the form of             Ax²  +  Bx  +  C  = 0    ?

             Yes !                    A=3      B=5   C=2

             so you can use the quadratic formula to solve it.

            x =  [ -B plus or minus √(B²-4AC) ] / 2A

               =  [ -5 plus or minus √(5² - 4·3·2) ] / 2·3

               =  [ -5 plus or minus √(25 - 24)  ]  /  6

               =  [  -5 plus or minus  √1          ]  /  6

           x  =        -4 / 6  =  -2/3
and
           x  =        -6 / 6  =  -1 .    
_______________________________________

The second question says

                                         4x²  +  5x  -  1 = 0

Is this in the form of          Ax²  +  Bx + C = 0  ?

           Yes it is !            A=4     B=5   C= -1  

         so you can use the quadratic formula to solve it.

            x =  [ -B plus or minus √(B²-4AC) ] / 2A

Now, you take it from here.

6 0
3 years ago
HELPPP!!!!!
shtirl [24]
42 is the LCD of all the above fractions.

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