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Mathematics: Post your doubts here!

Rutzaba

Rutzaba

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saqib.tariq said:
how the gradient of AC is -1/2 plz help :(View attachment 42725
Click to expand...
this is p1 isnt it?
 
B

Browny

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upload_2014-5-16_22-7-34.png
Hello, do you know,
-how to use mid points and frequencies to find the mean in above cumulative frequency graph?
-how to find sd (standard deviation)for A and B ?
upload_2014-5-16_22-9-1.png Marking scheme answer
 

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Rutzaba

Rutzaba

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daredevil said:
ummm do it ^__^
thanks in advance ^__^
Click to expand...
seeu got the value of t as 1.2765
t= tanx
1.2765 =tan x
x= tan inverse of 1.2765
one ans is 0.905
let this be alpha
tan is +ve in the first and the third quadrant
in the third quadrant pi +alpha
=4.04 radians but the range says tha answer shudnt be greater than pi
so 4.04 minus 2pi imagine the line having gone back past the 2nd quad the frst quad then the fourth quad then again it comes to the third quad a whole circle= 360degrees = 2pi
thats y we minused
4.04- 2pi =-2.24 rads
 
Rutzaba

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A star said:
Rutzaba can you please solve Q9 (ii) same paper :D
Click to expand...
1 - (√2)i was a second root of the equation. So you now have
p(x) = (x - 1 -√2 i)(x - 1 + √2 i) q(x)

p(x) = (x^2 - 2x + 3) q(x)

where q(x) is now a quadratic polynomial.

Just use long division to divide p(x) by (x^2 - 2x + 3)

You should find
q(x) = x^2 + 2x + 2

Use the quadratic formula to find the remaining two roots.
 
A

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Rutzaba said:
1 - (√2)i was a second root of the equation. So you now have
p(x) = (x - 1 -√2 i)(x - 1 + √2 i) q(x)

p(x) = (x^2 - 2x + 3) q(x)

where q(x) is now a quadratic polynomial.

Just use long division to divide p(x) by (x^2 - 2x + 3)

You should find
q(x) = x^2 + 2x + 2

Use the quadratic formula to find the remaining two roots.
Click to expand...
do you remember it from A levels or just seem babyish due to uni math ?
 
Rutzaba

Rutzaba

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i dunno wat u think of uni maths as... but its extremyly rubish a right pain in the.... it has rusted my maths badly and half the formulas in intermediate we dint even have in our course
 
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