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junior college 1 | H2 Maths
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hello pls help me with these questions? Thank u!!
Date Posted: 4 years ago
Views: 336
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Q7
For part two, observe that gradient is 0 when t = 1 and gradient = infinity when t is too positive small.
The thing is, the denominator t * e^t is already a small fixed number of e when t = 1, and it gets even smaller when t is lowered to decimals, whereas the numerator "increases in magnitude" from 0 to 1 as t goes from 1 to 9 (the negative sign is for downsloping purposes only).
With the magnitude of the numerator increasing to 1 and the denominator decreasing to 0, it is why the tangent becomes steeper and steeper as t goes to zero.
For part two, observe that gradient is 0 when t = 1 and gradient = infinity when t is too positive small.
The thing is, the denominator t * e^t is already a small fixed number of e when t = 1, and it gets even smaller when t is lowered to decimals, whereas the numerator "increases in magnitude" from 0 to 1 as t goes from 1 to 9 (the negative sign is for downsloping purposes only).
With the magnitude of the numerator increasing to 1 and the denominator decreasing to 0, it is why the tangent becomes steeper and steeper as t goes to zero.
Date Posted:
4 years ago
Thank u!!
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An idea for Q8, which is quite messy. Let me know if you do not understand what I wrote, need more explanation, my workings are too messy etc.
Date Posted:
4 years ago
may i know how u managed to get the graph for 8 ( iii) ?
I plotted this on Desmos parametric calculator using the coordinates (-2a cot t, 2a sin^2 t) and played around with the values of a from 0 to 10.
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