heelllppp pls binomial q (1 Viewer)

yashbb

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idk what i have done wrong, but there are no worked solutions to this problem.
1632277954571.png
how do i do it.
(srry for blurry image)
 

CM_Tutor

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You can also do this with the whole expression in sigma notation. Since , we can say:


The only term in in this expansion will occur when .

So, the required term is:

 

Lith_30

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Can you help with this pls
I've tried to solve it, but I haven't learnt any Ext 2 level projectile motion stuff yet. I'll try to learn it over the next few weeks if possible.
 

username_2

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Can you help with this pls

20210920_135959.jpg
So i remember doing this question and I couldn't figure it out then but I have now. I have put the solution down below. Anything you didn't understand, shoot a reply to this... here goes.

Drawing-10.sketchpad.png20210922_222122.jpg20210922_222205.jpg

Ok so I would even think that many people will get up to this point. But here's the little trick - the keyword in the question is show:

You don't have to solve the equation to show that k = 0.158, you can just substitute in the value for LHS and RHS and check if this is a valid solution to the equation. (Btw there is no good manual way to solve this equation other than precise graphing but that is impossible in an exam). Hence, is my solution - ie the next steps - below. (This part might not be the best but do give suggestions about any other methods you might know)

20210922_222231.jpg

Hope that is a bit more helpful! :)
 

CM_Tutor

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Looking at @username_2's equation 1, and setting , we have:


The solutions are thus the intersection of a diagonal line through the origin and the absolute value of a hyperbola that has as its two asymptotes the -axis and . There will likely be three solutions, one with , and likely two between . We know we aren't looking for the solution above 0.2379..., so must be in the region where , and so can solve:


The other solution, above 0.2379..., is in the region where , and so can solve:


So, the solution certainly can be found by HSC methods. There are indeed three solutions. And, either I have made an algebraic mistake(s) or there is a flaw in equation 1, as I haven't got .
 

username_2

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Looking at @username_2's equation 1, and setting , we have:


The solutions are thus the intersection of a diagonal line through the origin and the absolute value of a hyperbola that has as its two asymptotes the -axis and . There will likely be three solutions, one with , and likely two between . We know we aren't looking for the solution above 0.2379..., so must be in the region where , and so can solve:


The other solution, above 0.2379..., is in the region where , and so can solve:


So, the solution certainly can be found by HSC methods. There are indeed three solutions. And, either I have made an algebraic mistake(s) or there is a flaw in equation 1, as I haven't got .
Hey... um its not m abs(B/(B-18k)) = k/m, its ln(abs(B/(B-18k))) = k/m. Sorry for the terrible handwriting
 

CM_Tutor

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Hey... um its not m abs(B/(B-18k)) = k/m, its ln(abs(B/(B-18k))) = k/m. Sorry for the terrible handwriting
*That* explains my confusion!

In that case, I agree that there is no reasonable way to solve the equation.

Though, for MX2, if we want to be rigorous, you might look at


and note that it is defined for and continuous so long as , and then show that and have opposite signs to conclude that there must be at least one root on and thus there must be a root at (to 3 significant figures).
 

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