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anonymous
 5 years ago
Work cycle begins by typing on the body of doc for 30 minutes at 35wpm then takes a 5min break and works on ref document for 12 min at 20 wpm. sketch a graph of typing rate R(t) as a function of t during one typing
anonymous
 5 years ago
Work cycle begins by typing on the body of doc for 30 minutes at 35wpm then takes a 5min break and works on ref document for 12 min at 20 wpm. sketch a graph of typing rate R(t) as a function of t during one typing

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have (30, 35) and (12, 20) so my function is R(t)=5/6t + 10...do I need to include 5 minute breaks somewhere

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then I have to sketch a graph of total number T(t) of words that are typed as a function of time t during one typing cycle...I need several equations with different domains for over next 5 minutes...I'm stuck

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0can the f(t) be piece wise?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would imagine bc I will have several line segments that make up a polynomial line graph

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0 35 ; t=[0,30] f(t) =  0 ; t=(30,35)  20 ; t=[35,47]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the given coordinate axis has x values from 0 to 60 and yaxis 01500

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0need diagonal lines, right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I would say no. a graph can be disjoint, thre are no rules against that that I am aware of.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0for 35 mins the speed was constant, then dropped to 0 then picked up at 20 at the time intervals indicated is the best I can make out of the problem :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks I will keep working
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