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tyPut( x^2 + y^2 = 1 )back in the original equation they gav and write y in terms of x. solve for x then find y.
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tyPut( x^2 + y^2 = 1 )back in the original equation they gav and write y in terms of x. solve for x then find y.
The whole qn? or from( x^2 + y^2 = 1)do it for me plz :/
from( x^2 + y^2 = 1)The whole qn? or from( x^2 + y^2 = 1)
ohh okso take (x^2 + y^2 =1 ) and put it in the equation , you'll get
1 = 2 ( x^2 - y^2)
and since y^2= 1-x^2
1= 2( x^2 - (1- x^2)) expand this
4x^2 = 3
x^2 = 3/4
x = square root of (3/4)
y^2 = 1-x^2
substitute x in the equation
y= 0.5
Haha no problem.ohh ok
brain stopped working
thank u sooo much! ^_^
RoOkaYya Gplz can someone help with mechanics question on resloving! its been giving me a lot a issues...
heres the link:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_41.pdf
it is question number 3 roman numeral 2.
thx.
GOOD LUck then.Nope, but I will appear for that in June session, so will solve it in April ^_^
bludlynsolja
i)help needed urgently
View attachment 48023
I just solved that. :/i)
equate both equations (solve simultaneously)
2x^5 + 3x^3 = 2x
2x^5 + 3x^3 - 2x =0
x ( 2x^4 + 3x^3 - 2 ) = 0
x=0 and ( 2x^4 + 3x^3 - 2 ) = 0
( 2x^4 + 3x^3 - 2 ) = 0 [HENCE SHOWN]
ii)
let x^2 = y
2y^2 + 3y -2 = 0
solver for y
y= x^2
replace for the values u got as y
solve for x
where"I just solved that. :/
Book. Where else?where"
hihi okiiBook. Where else?
thanx u solved it on time have test tomorrowi)
equate both equations (solve simultaneously)
2x^5 + 3x^3 = 2x
2x^5 + 3x^3 - 2x =0
x ( 2x^4 + 3x^3 - 2 ) = 0
x=0 and ( 2x^4 + 3x^3 - 2 ) = 0
( 2x^4 + 3x^3 - 2 ) = 0 [HENCE SHOWN]
ii)
let x^2 = y
2y^2 + 3y -2 = 0
solver for y
y= x^2
replace for the values u got as y
solve for x
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