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The first three terms would mean the constant + the x + the x2. But if any of them is 0, then we must replace them by x3. And so on.
In other words, the first three terms refer to the first three NON-ZERO terms.
We do not need to pick out anything in x3 if the constant, the x and the x2 are not zero at all.
In other words, the first three terms refer to the first three NON-ZERO terms.
We do not need to pick out anything in x3 if the constant, the x and the x2 are not zero at all.
Date Posted:
5 years ago
Correction to 13a!
I made a mistake.
We need to add one more term so (5 3) 1^2 (x - 2x2)^3 +...
so we need to add another 10 (x^3) +...
so the final answer is 1 + 5x - 30x3 + ...
This is because I had not anticipated that the term in x2 is 0, so I need to include all the x3.
I made a mistake.
We need to add one more term so (5 3) 1^2 (x - 2x2)^3 +...
so we need to add another 10 (x^3) +...
so the final answer is 1 + 5x - 30x3 + ...
This is because I had not anticipated that the term in x2 is 0, so I need to include all the x3.
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For Q15 (in the earlier post), we only need terms up to x3 if the constant, the x, the x2 and the x3 are all non-zero. We therefore do not need to expand the (2x - 3x2)3 fully or it will be very very long.
Date Posted:
5 years ago