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# Some C4 easy questions watch

1. I got stuck on some of these questions
For question 13

I found t = 0, -2 and 2 on part a
for part c, I integrate part c and get (2t^5)/5 - (8t^3)/3 = f(x)

Then I subs t = 2 and t = -2 into f(x)
then take f(2) - f(-2) and get the ans -17/1/15? Is this the right ans?

For question 6, I dont know how to find value of x after getting all the 4 terms?

Thanks
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2. (Original post by prettyboy_bn)
For question 6, I dont know how to find value of x after getting all the 4 terms?

solve that and substitute it in
3. 1-3x=97

solve that and substitute it in
Do u mean x = (1 - 97)/3 ?
4. Anyone?
5. (Original post by prettyboy_bn)
Do u mean x = (1 - 97)/3 ?
no - you've just expanded so if you find the solution to and substitute it into your expansion, you'll find an approximation to
6. no - you've just expanded (1-3x)^{\frac{3}{2}} so if you find the solution to 1-3x=97 and substitute it into your expansion, you'll find an approximation to 97^{\frac{3}{2}}
ok so the solution to is and I subs it into the expansion (4 terms) and I got - 51,695 which is the approximation to ?
7. (Original post by prettyboy_bn)
ok so the solution to is and I subs it into the expansion (4 terms) and I got - 51,695
8. ergo, your expansion is incorrect
I just checked with the answer booklet, I got the expansion right. Their answer for x = 0.01?

Anything for question 13?

Thanks
9. (Original post by prettyboy_bn)
I just checked with the answer booklet, I got the expansion right. Their answer for x = 0.01?

Anything for question 13?

Thanks
letting gives an approximation to , not
10. For question 13 dont you intergrate within t = 0 and t=2?
O.o I could be wrong though
11. (Original post by Snapshot13)
For question 13 dont you intergrate within t = 0 and t=2?
O.o I could be wrong though
At t=0, the point is (1,0), at t=2 the point is (-3,0). The issue is, if you integrate from t=0 to t=2, you don't get the area of the full graph.

So, the OP is correct integrating from t=-2 to t=2

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