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AS Maths - Write the following as a power of 4

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Reply 20
Avatar for suggsina
suggsina
9
Original post by House Dagoth
it says a power of 4.

so 3^4, 5^4, 10^4 etc.

EDIT: oops, did I get it the wrong way round? :facepalm: sorry OP. just came back from a long day


lol we all know the feeling.
Reply 21
Avatar for MissBrown
MissBrown
1
Original post by funkydee
it explains the logic and provides working out.


For more involved functions maybe but if students are resorting to this in the first week of AS maths then they will not in the main develop any understanding.

If you are telling people to use wolfram to find 4^3 then Im sorry your not giving good advice IMO. This is something a average GCSE student should be able to deduce.

My comments are in no way related to the OP.
Reply 22
Avatar for JoeUtd
JoeUtd
OP
Another question..

The points A, B and C have co-ordinates (1,2,3) (-1,2,8) (3,-2,p) respectively, where p is an interger.

i) Show that AB^2 = 29

ii) Find AC^2 in terms of p

The question is slightly confusing. Does it mean A's coordinates are 1, -1 and 3 as that's the first digit of every bracket? Or is A the entire first bracket?
Original post by JoeUtd
Another question..

The points A, B and C have co-ordinates (1,2,3) (-1,2,8) (3,-2,p) respectively, where p is an interger.

i) Show that AB^2 = 29

ii) Find AC^2 in terms of p

The question is slightly confusing. Does it mean A's coordinates are 1, -1 and 3 as that's the first digit of every bracket? Or is A the entire first bracket?


Entire first bracket!

Hopefully you know how to apply Pythagoras' Theorem in 3D?
Reply 24
Avatar for JoeUtd
JoeUtd
OP
Original post by Mr M
Entire first bracket!

Hopefully you know how to apply Pythagoras' Theorem in 3D?


Haven't covered it yet. I'll look it up. Could you explain the basics of it?
Original post by JoeUtd
Haven't covered it yet. I'll look it up. Could you explain the basics of it?


Well you know in 2D that c2=a2+b2c^2 = a^2 + b^2

In 3D, d2=a2+b2+c2d^2 = a^2 + b^2 +c^2

In 2D, the length of a line segment is (x2−x1)2+(y2−y1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

In 3D, the length of a line segment is (x2−x1)2+(y2−y1)2+(z2−z1)2\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

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