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srry. i hate functions and complex numbers and this range and domain thingy :/
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srry. i hate functions and complex numbers and this range and domain thingy :/
Even I hate functionssrry. i hate functions and complex numbers and this range and domain thingy :/
Q10)Need help in p1 O/N/06 q:10 (iii) (iv) (v) and M/J/07 q:11 (ii) (iv)
and izzah saved the day!!!!Q10)
c) The answer to part two is f(x)= (x - 1.5)^2 - 2.25
we know that it will be graphed as a parabola that opens upwards (because x has a positive coefficient).
So we need too find the minimum value of f(x) ie -2.25. You can visualize that all the values of f(x) are greater than or equal to -2.25 since it is the minimum point of the curve.
iv) Since the graph is a parabola, every "y" value has 2 corresponding x values, ie the graph is not a 1:1 function. Only 1:1 functions have an inverse so this graph doesn't.
v) let 'radical x' be represented by z.
so x = z^2
z^2 - 3z - 10 = 0
solve the quadratic equation and you'll get z=-2 and z= 5
replace z with 'radical x'. Since anything squared is always positive, -2 is discarded. Now 'radical x' equals 5. square both sides. Final answer: x= 25
and izzah saved the day!!!!
ps. i love ur name
Thank uQ10)
c) The answer to part two is f(x)= (x - 1.5)^2 - 2.25
we know that it will be graphed as a parabola that opens upwards (because x has a positive coefficient).
So we need too find the minimum value of f(x) ie -2.25. You can visualize that all the values of f(x) are greater than or equal to -2.25 since it is the minimum point of the curve.
iv) Since the graph is a parabola, every "y" value has 2 corresponding x values, ie the graph is not a 1:1 function. Only 1:1 functions have an inverse so this graph doesn't.
v) let 'radical x' be represented by z.
so x = z^2
z^2 - 3z - 10 = 0
solve the quadratic equation and you'll get z=-2 and z= 5
replace z with 'radical x'. Since anything squared is always positive, -2 is discarded. Now 'radical x' equals 5. square both sides. Final answer: x= 25
Got a question guys, please help:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3)
then ul get 1 equation by putting it in the distance formula... and the othr equation by putting it in the gradient formula...Yar frst u take lcm of this equation
That will give you xb + ay = ab
Then at p where on x axis y = 0 if u put that value of y into abv equ ul get that a=x
then at Q y axis x wud be zero... put that into eq abv ul get y=b
P (a,0) and Q ( 0, b)
Lol.arghh functions!!! id confuse you more if id reply... this is a job for Binyamine
Got a question guys, please help:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_13.pdf
Q3)
may i be darned if i eva get that neat and clean !Rutzaba showed you the way.
( x / a ) + ( y / b ) = 1
Use the first information; to find P and Q.
On x-axis, y=0
x = a. Hence Coordinate P ( a , 0 )
On y-axis, x = 0
y = b. Hence Coordinate Q ( 0, b)
Next nformation; Length PQ is (45^ (1/2) )
manipulating the data;
a^ 2 + b ^ 2 = 45..............................First Equation
Next Information; Gradient = -1/2
b / (-a) = -1/2
a = 2b .............................................Second Equation
Solving the First and Second Equation Simultaneously would yield the following answer
b = 3 ; a = 6
may i be darned if i eva get that neat and clean !
well not really. i got the dirtiest wrk inda wrldLol, it is not that neat and clean. You are the girl and it is thought that girl do a cleaner and more neat work.
Yes its correct, my teacher gave it to me as a home assignment.are u sure this question is ryt ? :S
Need help in p1 O/N/06 q:10 (iii) (iv) (v) and M/J/07 q:11 (ii) (iv)
Need help in p1 O/N/06 q:10 (iii) (iv) (v) and M/J/07 q:11 (ii) (iv)
Yar mujh se nhi horha :/Need help in this question plz.
given that sinQ - cosQ /sinQ + cosQ= 6sinQ/cosQ
Find the exact values of tanQ
Need help in this question plz.
given that sinQ - cosQ /sinQ + cosQ= 6sinQ/cosQ
Find the exact values of tanQ
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