O-Level Additional Math

[SKT]

CoffeeCat:

SKT:

Hi,


LCM of A&B is 4x², B&C is 12x³, A&C is 6x³. Find all possible expressions C.

lol this is like a deductive puzzle.
Remember the concept of lcm is about choosing the highest index of prime factor.
lcm (A & B) = 4 x ^2 --- [1]
lcm (B & C) = 12 x^3 --- [2]
lcm (A & C) = 6 x^3 --- [3]

From [1] and any of [2] or [3] you can tell C must have x^3.
From [3] & [1] you can tell A has only factor 2 and not 4. B is the one with 4.
From [3] & [2] you can tell C must have factor 3 and possibly 2 but not 4.
Therefore possible expressions of C are
x^3, 3x^3, 6x^3.

Thank you CoffeCat.

[OK Lor]

Hi Sir,


What is the least possible value of
(a^2 + 9)^(1/2) + [(b-a)^2 + 4 ]^(1/2) + [(8-b)^2 + 16]^(1/2)
for real numbers a and b?

Thanks.

[liketoeat]

Hi Guanhui,


Please help me with the following:

If B = {a, {a}, b, {b}, {c}, d}, state whether each of the following is true or false.

(a) {a} is a subset of B

(b) {{a}} is a subset of B

{a, {a,c}, d} is a subset of B

Thank you

[k1ndan]

What is the last digit for 3^2010?


^: to the power of

Thanks

[fromnuaa]

answer: 1


3x3=9
9x9=81
1x1=1

k1ndan:

What is the last digit for 3^2010?

^: to the power of

Thanks

[k1ndan]

fromnuaa:

answer: 1


3x3=9
9x9=81
1x1=1

k1ndan:

What is the last digit for 3^2010?

^: to the power of

Thanks

Hi,

Thanks for the answer, but can you elaborate more on how you get the answer. Look complicated to me.

Thanks.

[fromnuaa]

sorry answer should be 9


2010 mod 4 =2

3x3=9

k1ndan:

fromnuaa:

answer: 1

3x3=9
9x9=81
1x1=1

[quote=\"k1ndan\"]What is the last digit for 3^2010?

^: to the power of

Thanks

Hi,

Thanks for the answer, but can you elaborate more on how you get the answer. Look complicated to me.

Thanks.[/quote]

[k1ndan]

Hi all,


Please help to solve the followin Sec 1 Maths Questions:

Q1) 3,12,25,42 are the 1st four terms of a number sequence.

(a) What is the nth term of the sequence? Express in terms of n.

(b) What is 40th term?

(c) Which term is the number 1537?



Q2) Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 8 = 2 x 4^2

(a) Express the sum of S in terms of n in the nth line sequence.

Thanks.

[k1ndan]

Hi all,


Please help to solve the followin Sec 1 Maths Questions:

Q1) 3,12,25,42 are the 1st four terms of a number sequence.

(a) What is the nth term of the sequence? Express in terms of n.

(b) What is 40th term?

Which term is the number 1537?



Q2) Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 8 = 2 x 4^2

(a) Express the sum of S in terms of n in the nth line sequence.

Thanks.

[Dharma]

k1ndan:

Hi all,


Please help to solve the followin Sec 1 Maths Questions:

Q1) 3,12,25,42 are the 1st four terms of a number sequence.

(a) What is the nth term of the sequence? Express in terms of n.

(b) What is 40th term?

(c) Which term is the number 1537?



Q2) Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 8 = 2 x 4^2

(a) Express the sum of S in terms of n in the nth line sequence.

Thanks.

Q1) 3,12,25,42 are the 1st four terms of a number sequence.

(a) What is the nth term of the sequence? Express in terms of n.
T1 = 3
T2 = 3 + (5 + 4) = 3 + 5(1) + 4(1)
T3 = 3 + (5 + 4) + (5 + 4 + 4) = 3 + 5(2) + 4(1 + 2)
T4 = 3 + (5 + 4) + (5 + 4 + 4) + (5 + 4 + 4 +4) = 3 + 5(3) + 4(1 + 2 + 3)

Tn = 3 + 5(n-1) + 4(n-1)(n)/2 = 3 + 5n – 5 + 2n^2 – 2n = 2n^2 + 3n - 2


(b) What is 40th term?

T40 = 2(40)^2 + 3(40) – 2 = 4918

(c) Which term is the number 1537?

2n^2 + 3n – 2 = 1537
2n^2 + 3n – 1539 = 0
(2n + 57)(n – 27) = 0
2n + 57 = 0 or n – 27 = 0
n = -57/2 (rejected) or n = 27
27th term => 1537


Q2)
Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 32 = 2 x 4^2

(a) Express the sum of S in terms of n in the nth line sequence.

Sn = 2(n + 1)^2