treefingers
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Either way, 2 doesn't equal 1 haha 
Secondary Maths questions, give me your maths questions
- Thread starter Nasma
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- #152
sorry, was a bit sloppy by me. should have explained that what I meant is that if the numerator is zero then the fallacy is obvious, and so the example is pointless. i.e. you only divide by (x-y) because you don't immediately realise x-y=0...if you did, you wouldn't do it.
treefingers
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On an Argand diagram, O is the origin and P represents the point 3 + i. The point Q
represents a + bi, where a and b are both positive. If triangle OPQ is equilateral, find a
and b.
I've tried using vectors, locus, cis etc but can't seem to get anywhere.
represents a + bi, where a and b are both positive. If triangle OPQ is equilateral, find a
and b.
I've tried using vectors, locus, cis etc but can't seem to get anywhere.
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- #154
Lots of ways to do this, but the most straight-forward is just to use plain old geometry:
- Draw your diagram.
- Find the angle between the Real axis and OP (let's call this x Radians).
- The angle between the Real axis and OQ is then Pi/3 + x. Draw in OQ, and hopefully you can take it from here.
treefingers
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- #157
sin^3(x^2) is just the standard notation for (sin(x^2))^3. Let u = sin(x^2) and off you go.
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- #159
It's been a while, but if you post a couple of them I'll see if I can help.
Simplify:
sin2x
cos(2x)+1
sin2x
cos(2x)+1
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- #163
take a look at your trig identities, and think about what you could do to the denominator to make it a single term (think multiplication).
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- #164
Q20 on combinations if anyone can solve it
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- #165
Q20 on combinations if anyone can solve it
Consider which combinations of 3 letters (vertices) will not form a triangle, and find the total number of these. Call this X.
Then to solve your problem, all you need is to find all possible 3-letter combinations of A-I and subtract X.
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- #166
Spent a day thinking about the question eventually got it
a) 5C2 x 4C1 + 5 C1 x 4C2 = 70 (1st row choose 2 and 1 from the 2nd row) + (1 from the 1st row and 2 from the 2nd row)
b) If A was a must point ( 1C1 x 4C1 (1st row) x 4C1 (2nd row)) + (1C1 x 4C0 (1st row) x 4C2 (2nd row))= 22
a) 5C2 x 4C1 + 5 C1 x 4C2 = 70 (1st row choose 2 and 1 from the 2nd row) + (1 from the 1st row and 2 from the 2nd row)
b) If A was a must point ( 1C1 x 4C1 (1st row) x 4C1 (2nd row)) + (1C1 x 4C0 (1st row) x 4C2 (2nd row))= 22
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- #167
Question 19 , the answer is 13.4 km
Can someone explain the solution for me?
Question 19 , the answer is 13.4 km
Can someone explain the solution for me?
Draw a diagram like this:
Then use the cosine rule to work out the value of x (13.4).
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- #169
Draw a diagram like this:
Then use the cosine rule to work out the value of x (13.4).
Thanks buddy, that helped solve the problem. I was reading the question wrongly for this one
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- #170
Got another tricky question. Q 36
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- #171
The answer for 36a) 1.77km 36 b) 1.17 km
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- #172
Internal angles of the triangle are 40, 25, 115
from there use the sine rule. Try it yourself then I will post a solution if you are having any problems .
from there use the sine rule. Try it yourself then I will post a solution if you are having any problems .
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- #173
Im still a bit confused, not sure if I drew the diagram correctly. Would be good if someone could have a look for me.
Uploaded with ImageShack.us
Uploaded with ImageShack.us
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- #174
This is how I drew it.
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