Please help me w this!!
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<h3>Answer:</h3>
One of ...
- 6,301.005
- 6.301005×10³
It depends on where you are.
In the US, standard form is the way you have written the number. Your number would be 6,301.005.
The attachment gives some guidance for number formats in American publications. In certain publications, thin spaces are used instead of commas to separate groups of digits. The comma (or space) is often optional, but can enhance readability for large numbers.
In Britain, standard form is the form called "scientific notation" in the US. Your number would be 6.301005×10³.
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.