MaxWong

First we figure out what's (10^27 - 1800) mod 9

(10^27 - 1800) mod 9

= 10^27 mod 9 - 1800 mod 9

= 10^27 mod 9

Then we find sequence for 10^n mod 9

10^1 mod 9 = 1

10^2 mod 9 = 1

10^3 mod 9 = 1

10^4 mod 9 = 1

Therefore 10^27 mod 9 = 1

Substitute (10^27 - 1800) mod 9 = 1 into the eq.

1 - u ≡ 1 mod 9

u = 0

They chirp together every 24 seconds.

First cricket: 0 6 12 18 24

Second cricket: 0 8 16 24

Thanks Alan :D.

a subtext 1 = 7/7 = 1

a subtext 2 = 8/14 = 4/7

a subtext 3 = 9/33 = 3/11

a subtext 4 = 10/70 = 1/7

a subtext 5 = 11/131

a subtext 6 = 12/222 = 2/37

As it goes nearer and nearer to 0,

     lim        a subtext n = a subtext infinity = 0.

x-> infinity

Answer is d.

Square the second equation.

(x+y)^2=9

x^2 + 2xy + y^2 = 9

Substitute the first equation to this.

2xy + 9 = 9

2xy = 0

xy = 0

First let x = 0

0+y = 3

y = 3

A solution is (0, 3)

Let y = 0

x + 0 = 3

x = 3

Another solution is (3, 0)

Therefore there are 2 solutions, namely (0, 3) and (3, 0)

If a = -2i + 3j - 6k

Does the vector go from (0, 0, 0) to (-2, 3, -6)?

For the second question:

Square the 2nd equation:

(x+y)^2=16

x^2 + 2xy + y^2 = 16

Substitute x^2 + y^2 = 8

2xy + 8 = 16

2xy = 8

xy = 4

y = 4/x

Substitute this into the 2nd equation:

x + 4/x = 4

x^2 - 4x + 4 = 0

(x-2)^2=0

x - 2 = 0

x = 2

y = 4/2 = 2

For the 1st question:

Square the 1st equation.

(x+y)^2=256

x^2 + 2xy + y^2=256

Substitute x^2 + y^2 = 128

2xy + 128 = 256 <---- That goes back from second-order to first-order.

2xy = 128

xy = 64

y = 64/x ---(3)

Substitute (3) into x+y = 16
x + 64/x = 16

x^2-16x + 64 = 0 <---- That goes back from first-order to second-order, funny.

(x-8)^2=0

x-8 = 0

x = 8

y = 64/8 = 8

Or if you really mean 1/2 * 4/3 * pi * 253 = 2/3 * pi * 253, that is an identity.