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# some FP1 doubts watch

1. How does 2(2^k) + 6(6^k) simplify into 6(2^k + 6^k ) -4(2^k)?

Secondly, suppose that there are three matrices, A B and C , such that AC=B
If we were given B and C, and asked to find A, would it be A inverse * B or B * A inverse?

And lastly, how do I prove that the line 2y-4x-1=0 is a tangent to the curve?
(The parametric equations of the curve are x=(t^2) , y=2t and the equation of a tangent at the point with parameter t is ty = x+(t^2))

thanks!
2. 2(2^k) + 6(6^k) = 2(2^k) + 6(6^k) +4(2^k) - 4(2^k)

then collect terms.
3. (Original post by confuzzled92)
How does 2(2^k) + 6(6^k) simplify into 6(2^k + 6^k ) -4(2^k)?
Write and expand the bracket, then take 6 out as a common factor of the corresponding two terms.

(Original post by confuzzled92)
Secondly, suppose that there are three matrices, A B and C , such that AC=B
If we were given B and C, and asked to find A, would it be A inverse * B or B * A inverse?
Matrix multiplication depends on whether you're multiplying on the left or the right. What do you need to multiply AC by to get rid of A? Which side do you need to multiply on? You need to do the same, to the same side, for B (because B=AC).

(Original post by confuzzled92)
And lastly, how do I prove that the line 2y-4x-1=0 is a tangent to the curve?
(The parametric equations of the curve are x=(t^2) , y=2t and the equation of a tangent at the point with parameter t is ty = x+(t^2))
You have to show that there is a point on the line 2y-4x-1=0 which lies on the curve and that, at that point, the gradient of the two curves are the same.

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