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secondary 4 | E Maths
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please help..
Date Posted: 3 years ago
Views: 493
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your answer is correct time is 5 secs. but can you explain how
h is 126 +p and how t is 5 sec
h is 126 +p and how t is 5 sec
can you please explain how h is 126+ p and how t is 5 secs
ya got it. thanks
for maximum/minimum points u can just let whatever is in the brackets = 0 and solve. in this case u had (t-5) thats why t=5
alternatively you can use the method of how MAXIMUM point is (-b/2a).
in this equation,
h = -5t^2 + 50t + p + 1
so a = -5, b = 50, c = p+1
(look at the coefficients. what comes before "t^2" will be your a, what comes before "t" will be your b and what is left will be your c. *take note of your positive & negative signs!)
so maximum point would be
-50 / 2(-5) = -50 / -10 = 5
hope this helps!
alternatively you can use the method of how MAXIMUM point is (-b/2a).
in this equation,
h = -5t^2 + 50t + p + 1
so a = -5, b = 50, c = p+1
(look at the coefficients. what comes before "t^2" will be your a, what comes before "t" will be your b and what is left will be your c. *take note of your positive & negative signs!)
so maximum point would be
-50 / 2(-5) = -50 / -10 = 5
hope this helps!
Thank you so much for the detailed explanation. understood nicely.
done
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for quadratic equations of the form y=ax^2 + bx + c
use the formula x = -b/2a to find the x coordinate of the turning point.
use the formula x = -b/2a to find the x coordinate of the turning point.
Date Posted:
3 years ago
Thank you so much for explanation
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