O-Level Additional Math
-
Herbie:
note that 2s-3q=-(3q-2s)factorisation
Need help for this qn.
(6r-7t)(3q-2s)+(2s-3q)(3r-2t)
Tq
(6r-7t)(3q-2s)-(3q-2s)(3r-2t)
=(3q-2s)[(6r-7t)-(3r-2t)]
=(3q-2s)(6r-7t-3r+2t)
=(3q-2s)(3r-5t) -
(6r-7t)(3q-2s)+(2s-3q)(3r-2t)
=(6r-7t)(3q-2s)-(3q-2s)(3r-2t)
=(3q-2s)[(6r-7t)-(3r-2t)]
=(3q-2s)(6r-7t-3r+2t)
=(3q-2s)(3r-5t)
Rgds,
Concerndad -
Secondary Two Maths
Hi
Solve this equation
(x-2)^2 = 3(2x-3)(2-x)
(x-2)^2 means (x-2) to the power of 2.
Please help.
Thank you. -
suiyuan:
(x-2)^2 = 3(2x-3)(2-x)Secondary Two Maths
Hi
Solve this equation
(x-2)^2 = 3(2x-3)(2-x)
(x-2)^2 means (x-2) to the power of 2.
Please help.
Thank you.
(x-2)^2 = -3(2x-3)(x-2)
(x-2)^2 + 3(2x-3)(x-2) = 0 ( note: (2-x) = -(x-2)
(x-2)[(x-2) + 3(2x-3)] = 0
(x-2)[ x- 2 + 6x -9] = 0
(x-2)( 7x - 11) = 0
x- 2 = 0 or 7x - 11 = 0
x = 2 or 11/7 -
suiyuan:
(x-2)^2-3(2x-3)(2-x)=0Secondary Two Maths
Hi
Solve this equation
(x-2)^2 = 3(2x-3)(2-x)
(x-2)^2 means (x-2) to the power of 2.
Please help.
Thank you.
(x-2)^2+3(2x-3)(x-2)=0 (note that -(2-x)=+(x-2))
(x-2)[(x-2)+3(2x-3)]=0
(x-2)(7x-11)=0
x=2 or x=11/7 -
hometutors.sg:
Thank you for your time.
A good tutor / mentor would be able to select what is good for the student.JadeDry:
I currently use \"New Syllabus Mathematics\" (8th Grade) books, and would appreciate if you could recommend additional good quality publications.
Thanks in advance.
I have just finished:
Secondary one Mathematics Tutor: 1A and 1B.
Shinglee New Syllabus Mathematics 6th edition 1
and Singlee Mathematics Workbook 1.
I am homeschooled and have been studying singapore math from 1st grade, since my parents are my teachers I would like to see what you people think I should get for 8th Grade.
Thank you in advance for your assistance. -
Sec Two Maths
1) Given that x^2 = 5x – 1, find the value of 2x^4 + 2/x^4
2) a)Factorise 2x^10 – 13x^5 -15 completely.
b)Factorise 10x^10y – 65x^5y-75y
Note a) ^2 means to the power of 2 b) 2/x^4 means 2 divided by x^4
Please help.
Thank you -
suiyuan:
Do you happen to be from some IP school or sth?Sec Two Maths
1) Given that x^2 = 5x – 1, find the value of 2x^4 + 2/x^4
2) a)Factorise 2x^10 – 13x^5 -15 completely.
b)Factorise 10x^10y – 65x^5y-75y
Note a) ^2 means to the power of 2 b) 2/x^4 means 2 divided by x^4
Please help.
Thank you
Your teacher probably koped qns 1 from SMO.
Qns 1)
Lousy Method : solve for x, substitute and torture yourself in the process.
Proper Method : divide the quadratic equation by x throughout to get
x + 1/x = 5
You will get x^2 + 1/x^2 = 23 by squaring the above.
Repeat the process one more time.
Final ans should be 1054 if I have no calculation error.
Qns 2a)
Notice 10 is twice that of 5. hence imagine you substitute y=x^5. You will reduce the problem to factorising a quadratic expression.
Qns 2b)
Ambiguous.......Please use brackets.
10x^10y means 10x^(10y) or 10(x^10)y
Since it's part b, i will assume the latter.
First extract common factors from all terms, which is just 5y. After that what you have should be part a. -
Hello again, I have a problem that I am unsure about:
Simplify the following expression:
3/4((x squared)(y)) X 5/9((x)(z cubed)) / ((20/9)(x quadrupled)(y squared)(z))
These are my workings:
3/4((x squared)(y)) X 5/9((x)(z cubed))=
15/36((x cubed)(y)(z cubed))
15/36((x cubed)(y)(z cubed)) / ((20/9)(x quadrupled)(y squared)(z))=
3(Z squared) / 16(xy)
Are my calculations correct?
On a side note, is Z cubed/ Z= Z squared or Z cubed?
Thanks in advance. -
Sec Two Maths
Find the smallest possible value of the expression 3x^2 + 27y^2+5z^2-18xy-30z+125.
Note a) ^2 means to the power of 2
Please help.
Thank you
Hello! It looks like you're interested in this conversation, but you don't have an account yet.
Getting fed up of having to scroll through the same posts each visit? When you register for an account, you'll always come back to exactly where you were before, and choose to be notified of new replies (either via email, or push notification). You'll also be able to save bookmarks and upvote posts to show your appreciation to other community members.
With your input, this post could be even better 💗
Register Login