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secondary 4 | A Maths
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must i assume that abc are positive ? what if they happen to be negative ? will i get 2 sets of answers ?
Date Posted: 5 years ago
Views: 317
What is t? The number of hours after 0300?
i got a= -4 vs a=4. Then sketch . Then reject positive 4 becos it is supposed to be lower temp at the start .
am i right to say that it is best to sketch in step 1 ?
What's the value of c you obtained ? 30? If yes, then the temperature is 26 for both t = 300 and t = 1500, when a = -4
yes c=30 for both a= posi or neg 4. thanks !!
Then there's something odd about this question. When a = 4, both temperatures are 34°C. When a = -4, both temperatures are 26°C.
i just posted a pic . i used elimination method to eliminate positive 4
Yup I got your point. What I'm saying is b can't be π/12
why a has to be -4 :
if the graph starts from 0300, then it should be increasing (upward sloping) from 0300.
I.e the coefficient of cos(bt) should be positive. So -a has to be positive, which means a has to be negative
why a has to be -4 :
if the graph starts from 0300, then it should be increasing (upward sloping) from 0300.
I.e the coefficient of cos(bt) should be positive. So -a has to be positive, which means a has to be negative
i just posted pic about b
If you actually plotted the graph of T(t) = 4 cos(π/12 t) + 30 using a software (eg. Desmos), you will get T = 26 for both t = 300 and t = 1500. So this question is flawed
i posted pic of desmos . it looks like vertical value = 26 when horizontal = 0 and 24 hr .
Yup. So the graph can't just have b = π/12 without any translation.
Try plotting T(t) = -4(π/1200 t - π/4) + 30
Using b = π/12 implies that the temperature fluctuates too rapidly throughout the day
Try plotting T(t) = -4(π/1200 t - π/4) + 30
Using b = π/12 implies that the temperature fluctuates too rapidly throughout the day
i’m confused . what is wrong with y= 4cos (pi/12 x ) + 30 as per your previous previous comment ? it seems to fit the storyline of temp =26 , then 34 in middle , then back to 26 .
The problem is, your t starts from 0. It should start from 3. But the minimum temperature isnt at 12 midnight. It is at 3am.
The question says t is the time in hours. So at 3am, it would be 0300 hours, i.e t = 300. At 3pm, it would be 1500 hours, i.e t = 1500.
The question says t is the time in hours. So at 3am, it would be 0300 hours, i.e t = 300. At 3pm, it would be 1500 hours, i.e t = 1500.
is it the 3am refers to t=0 ? like the equation starts counting from 3am , means t=0 is the same thing as 3am ?
That's what the question was implying. But the definition of t is poorly worded. It should have been phrased as something along the lines of
't is the number of hours elapsed from 0300 hours of the present day'
't is the number of hours elapsed from 0300 hours of the present day'
i would interpret 3am as clock reading , not hours . Hours is the time lapsed from the clock
i agree the english is lacking . haha . will update this space when the student gets back to me with her sch explanation
thanks for helping although i’m not a bona fide student
No problem.
Anyway, the way t is equated in the parts of the question also don't make sense.
We never see things like 't = 0300' in Cambridge papers. t is variable of a value ,not a clock reading , similar to your mentioning of interpreting 3am as a reading and not hours.
Anyway, the way t is equated in the parts of the question also don't make sense.
We never see things like 't = 0300' in Cambridge papers. t is variable of a value ,not a clock reading , similar to your mentioning of interpreting 3am as a reading and not hours.
See 6 Answers
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this is actual qn
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answer for i
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to find b value
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5 years ago
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when t=0 and 24 , vertical value = 26
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sch answer part 1
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5 years ago
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sch ans part 2
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5 years ago
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