Y=1/4x−2y=−2x+3 help solve problem.
1 answer:
Answer:
(x, y) = (2 2/9, -1 4/9)
Step-by-step explanation:
Equate the values of y and solve for x.
1/4x -2 = -2x +3
(2 1/4)x = 5 . . . . . . . . add 2+2x to both sides
x = 20/9 = 2 2/9 . . . multiply by 4/9
y = -2(2 2/9) +3 = -4 4/9 +3 . . . . substitute for x in the second equation
y = -1 4/9
The solution is x = 2 2/9, y = -1 4/9.
You might be interested in
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.
.. 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.