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Let the origin of the coordinate system be the position of ship B at noon. Then the distance between the ships as a function of t (in hours) is
.. d = √((30 +23t)² +(17t)²)
.. = √(818t² +1380t +900)
The rate of change of distance with respect to time is
.. d' = (1/2)(2*818*t +1380)/√(818t² +1380t +900)
.. = (818t +690)/√(818t² +1380t +900)
At t=4, this is
.. d' = (818*4 +690)/√(818*16 +1380*4 +900)
.. = 3962/√19508
.. ≈ 28.37 . . . . . knots
The distance between the ships is increasing at about 28.37 knots at 4 pm.
It is close to 8. When you look for the square root of 59 you should get about 7.6811457... Round up and thats 8 round down that 7 but it should be 8.
Y = 2x - 2
x + 3y = 15
I'll do substitution by solving for y:
y = 2x - 2
Solve for y for the second equation:
y = 5 - x/3
Substitute:
2x - 2 = 5 - x/3
Add x/3 to both sides:
7x/3 - 2 = 5
Add 2:
7x/3 = 7
Multiply by 3:
7x = 21
Divide by 7:
x = 3
Substitute x in for one of the original equations:
y = 2(3) - 2
y = 6 - 2
y = 4
The solution set is (3, 4).