<h3>
Answer: (3, 0)</h3>
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Explanation:
Let's isolate x in the first equation.
x-2y = 3
x = 3+2y
Then we'll plug this into the second equation
Afterwards, solve for y.
2x + 4y = 6
2(3+2y)+4y = 6
6+4y+4y = 6
8y+6 = 6
8y = 6-6
8y = 0
y = 0/8
y = 0
Use this to find x.
x = 3+2y
x = 3+2(0)
x = 3
The solution is therefore (x,y) = (3, 0)
If you were to graph both lines, then they would intersect at the location (3,0).
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Checking the answer:
Plug x = 3 and y = 0 into the first equation.
x-2y = 3
3-2(0) = 3
3 - 0 = 3
3 = 3 that works
Repeat for the other equation
2x+4y = 6
2(3) + 4(0) = 6
6 + 0 = 6
6 = 6 that works as well
Both equations are true when (x,y) = (3,0).
The solution is confirmed.
Answer: The first option (A)
Step-by-step explanation:
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.
Step-by-step explanation:
It's 27/35, so 27/1 * 1/35