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secondary 4 | E Maths
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Can help me with qns 17, 19 and 20? Thank u
Date Posted: 5 years ago
Views: 581
Good afternoon Annela! I will look at the remaining questions later by 4 am if no one else has done them yet. If I can find graph paper, I may attempt the graph questions.
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Q20
Hope this helps! Impt parts in pink
Hope this helps! Impt parts in pink
Date Posted:
5 years ago
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Good evening Annela! Here are my workings for Q17a.
Date Posted:
5 years ago
Hi why is hpn3 =270 degrees?
Good evening Annela! Because P is due east of H, N1HP= 90 degrees.
This means that normal angle HPN3 is 180 - N1HP = 180 - 90 = 90, since both “norths” N1 and N3 are parallel to each other.
Then, reflex HPN3 = 360 - normal HPN3 = 360 - 90 = 270.
I will look at the rest later because I am on the move the whole day with students and other things.
This means that normal angle HPN3 is 180 - N1HP = 180 - 90 = 90, since both “norths” N1 and N3 are parallel to each other.
Then, reflex HPN3 = 360 - normal HPN3 = 360 - 90 = 270.
I will look at the rest later because I am on the move the whole day with students and other things.
The graph questions I will look at it again today but at a very late time, but I will need to find my own graph papers first; alternatively you can use Desmos online graphing calculator or data points plotter (for best fit curves and lines, suitable for linear law as well) to attempt the graph questions.
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Good evening Annela! Here are my workings for Q17b as a supplement to my previous workings for Q17a. I’m so sorry to have kept you waiting for the workings!
Note that my final answer in Q17a earlier should be in km, not in degrees, which I have written wrongly.
Note that my final answer in Q17a earlier should be in km, not in degrees, which I have written wrongly.
Date Posted:
5 years ago
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Good evening Annela! Here are my workings for Q19a.
Date Posted:
5 years ago
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Good evening Annela! Here are my workings for Q19b.
Date Posted:
5 years ago
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