Please someone factor 8 + 16 !!!!!
1 answer:
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Answer:
Step-by-step explanation:
Given
Required
How far did they row together
To do this, we simply add the distance they both covered.
This gives;
Take LCM
Simplify
<em>Hence, they both rowed 1/2 mile</em>
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.