Question

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I nedd help number 53
The key is
a) 100,120,100,80,100
b) 0.25s
c) 0.75s

Answer 1

Hi Nancy, well
a) Find the blood pressure at t=0, t=0.25, t=0.5, t=0.75, t=1.
As the pressure in each second is given by P(t) =100+20*sin(2*pi*t)
we have:
P(t=0)=100+20*sin(2*pi*0)=100+20*sin(0)=100+0=100
P(t=0.25)=100+20*sin(2*pi*0.25)=100+20*sin(pi/2)=100+20*1=120
P(t=0.5)=100+20*sin(2*pi*0.5)=100+20*sin(pi)=100+0=100
P(t=0.75)=100+20*sin(2*pi*0.75)=100+20*sin(3pi/2)=100+ 20*(-1)=80
P(t=1)=100+20*sin(2*pi*1)=100+20*sin(2pi)=100+ 0=100

b) as the range of the sine function is [-1,1], the term 20*sin(2*pi*t) reaches its maximum value when sin(2*pi*t)=1, that is when t=0.25 (because with t=0.25 we have sin(2*pi*0.25) = sin(pi/2) =1 that is the maximum)

c) thinking in the same way, we want t when sin(2*pi*t)=-1 and that is t=0.75 because sin(2*pi*0.75)=sin(3*pi/2)= -1

sorry for my bad English

Answer 2

Thank you for spend time do my home work I appciate, yoy are desver a medal

Answer 3

Can you help me one more I did post , go up about 3 or 4 more line

Answer 4

yeah, of course.

Answer 5

I will be there after I finish this question,ty

Answer 6

2pi=360??

Answer 7

100+(20sin360*0.25)?

Answer 8

yep

Answer 9

100+(20sin360*0.25) = 100 + 20sin(360/4) = 100+ 20 sin (90) = 100+ 20*1 =120