Mathematics: Post your doubts here!

[Lily9605]

do it for me plz :/

The whole qn? or from( x^2 + y^2 = 1)

 

[RoOkaYya G]

The whole qn? or from( x^2 + y^2 = 1)

from( x^2 + y^2 = 1)

 

[Lily9605]

so 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

 

[RoOkaYya G]

so 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

ohh ok
brain stopped working
thank u sooo much! ^_^

 

[Lily9605]

ohh ok
brain stopped working
thank u sooo much! ^_^

Haha no problem.

 

[bludlynsolja]

plz can someone help with mechanics question on resloving! its been giving me a lot a issues...
heres the link:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_41.pdf
it is question number 3 roman numeral 2.
thx.

 

[Thought blocker]

plz 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.

RoOkaYya G

 

[shaminou]

Nope, but I will appear for that in June session, so will solve it in April ^_^

GOOD LUck then.

 

[bludlynsolja]

oh so u doing mechanics next year?
best wishes!!

 

[bludlynsolja]

so anyone able to help with that question?

 

[Glory Hunter]

help needed urgently

 

[RoOkaYya G]

RoOkaYya G

bludlynsolja
use formula R^2= X^2 + Y^2

where X = horizontal components
Y= vertical component

solve for R
R= magnitude of of resultant force

for angle :

use formula tan theta= Y/X

theta= tan (inverse) Y/X

the direction will be anticlockwise from x plane (i direction of i,j,k position)

 

[RoOkaYya G]

help needed urgently
View attachment 48023

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

 

[Thought blocker]

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

I just solved that. :/

 

[RoOkaYya G]

I just solved that. :/

where"

 

[Thought blocker]

where"

Book. Where else?

 

[RoOkaYya G]

Book. Where else?

hihi okii

 

[Glory Hunter]

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

thanx u solved it on time have test tomorrow
thanx a lot

 

[RoOkaYya G]

thanx u solved it on time have test tomorrow
thanx a lot

welcome
whn u see big powers don get afraid
always try with disguised funtions

 

[Glory Hunter]

welcome
whn u see big powers don get afraid
always try with disguised funtions

lol ok thnx