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Cambridge 0606

Their answers are out of this world sometimes. Can't wait for IB.

Well, even textbooks have mistakes... :P

I got the same answer!

Book says x = pi/4

*facepalm*

We can just factorize it :)

4a^2 - 10a + 6 = 0

2a^2 - 5a + 3 = 0

(2a - something)(a - something)=0

We don't know what it is yet, but think of 2 numbers that satisfies:

\(-2x-y=-5\\ xy=3\)

We can easily see that x = 1 and y = 3.

(2a - 3)(a - 1) = 0

a = 3/2 or a = 1 :)

It's not the simplest form ._.

\(-6t \cdot 4m \cdot (-m)\\ =(-6\cdot 4)\cdot(t\cdot m\cdot(-m))\\ =-24(-tm^2)\\ =24tm^2\)

Simplest form should be 24tm^2...

\(\displaystyle\lim_{x\rightarrow 5}\)

LaTeX code: \displaystyle\lim_{x\rightarrow 5}

\(\sin(3t+\pi)+2\cos(3t)\\ =\sin(3t)\cos(\pi)+\cos(3t)\sin(\pi)+2\cos(3t)\\ =\sin(3t)\cdot(-1)+0+2\cos(3t)\\ =2\cos(3t)-\sin(3t)\)

\(x=\dfrac{\pi}{4}\\ \therefore y = \dfrac{1}{\sin\left(2\cdot\dfrac{\pi}{4}\right)}=1\\ \dfrac{dy}{dx}=(\csc(2x))'=-2\csc(2x)\cot(2x)\\ \dfrac{dy}{dx}|_{x=\frac{\pi}{4}}=-2(1)(0)=0\\ y-1=0(x-\dfrac{\pi}{4})\\ y=1\\ \therefore\text{The equation of the normal to }y=\dfrac{1}{\sin(2x)}\text{ when }x=\pi/4\text{ is }\boxed{y=1}\)

I think this is the first time I do an application question :P

Gonna solve that all :P

1)

\(x^2=x-1\\ x^2-x+1=0\\ \Delta = b^2 - 4ac = (-1)^2-4(1)(1)=-3\\ \therefore\text{No real solutions.}\)

2)

\(\sqrt{x+5}-2=6\\ \sqrt{x+5}=8\\ x+5=64\\ x=59\)

3)

\(\text{Note that:}\\\boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}\\ (\sqrt[3]{3}+\sqrt[3]{2})(\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9})\\ =(\sqrt[3]{3}+\sqrt[3]{2})\left((\sqrt[3]{2})^2-(\sqrt[3]{2})(\sqrt[3]{3})+(\sqrt[3]{3})^2\right)\\ =(\sqrt[3]{3})^3+(\sqrt[3]{2})^3\\ =3+2\\ =5\)

4)

\(x^4+3x^2+3=1\; |u=x^2\\ u^2+3u+2=0\\ (u+1)(u+2)=0\\ x^2+1=0\text{ or }x^2+2=0\\ x=i,x=-i,x=\sqrt{2}i,x=-\sqrt{2}i\)

5) 

\(\sqrt{10000}-\sqrt{81}+\sqrt{81}\\ =\sqrt{10000}\\ =100\)

6)

\(\dfrac{8}{x}=x+\dfrac{1}{x}\\ 8=x^2+1\\ x^2-7=0\\ x=\sqrt7\text{ or }x=-\sqrt7\)

7) Not enough information...

I tick it and my email address is all correct but still I can't receive it.

5^2*5^4*5^6*.......5^(2n) = (0.008)^(-30) find the value of n

\(\begin{array}{|rclrcl|} \hline 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& 0.008^{(-30)} \\ && 0.008 &=& \frac{8}{1000} \\ && &=& \frac{2^3}{10^3} \\ && &=& \left( \frac{2}{10} \right)^3 \\ && &=& \left( \frac{1}{5} \right)^3 \\ && &=& \frac{1}{5^3} \\ && &=& 5^{-3} \\ 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& (5^{-3})^{-30} \\ 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& 5^{(-3)\cdot (-30) } \\ 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& 5^{90} \\ 5^{2+4+6 \cdot \ldots \cdot 2n } &=& 5^{90} \\\\ 2+4+6 \cdot \ldots \cdot 2n &=& 90 \\ 2\cdot(1+2+3 + \ldots + n ) &=& 90 \\ && 1+2+3 + \ldots + n &=& \frac{(1+n)\cdot n}{2} \\ 2\cdot \Big(\frac{(1+n)\cdot n}{2} \Big) &=& 90 \\ (1+n)\cdot n &=& 90 \\ n^2+n -90 &=& 0 \\ (n-9)\cdot(n+10) &=& 0 \\ \hline \end{array}\)

\(n = 9 \text{ or } n = -10\)

Because n > 0:

n = 9

\(5^2\cdot 5^4 \cdot 5^6\cdot 5^8\cdot 5^{10}\cdot 5^{12} \cdot 5^{14}\cdot 5^{16}\cdot 5^{18} = 0.008^{(-30)} \)

= 2cos(3t) - sin(3t)

this is 8 * 8 :

Use the keyboard

\(8*8=64=8^2\)

Right?

Good job.

Let n be a positive integer and a be an integer such that a is its own inverse modulo n.
What is the remainder when a^2 is divided by n?

A modular inverse of an integer a (modulo n) is the integer \(a^{(-1)}\) such that
\(aa^{(-1)}=1 \pmod n\)

So let \(a^{(-1)} = a\), we have:

\(\begin{array}{|rcll|} \hline a\cdot a &=& 1 \pmod n \\ a^2 &=& 1 \pmod n \\ \hline \end{array}\)

The remainder when a² is divided by n ist 1

To figure out how much is 3% of 100,000,000, let's first understand the language of the problem. In mathematics, "of" is a word that generally refers to multiplication. Knowing this, the problem now changes from "3% of 100000000" to "3% * 100000000:"

Therefore, 3% of 100000000 is 3000000

The only possibilities for the third side of the triangle are  5, 4 and 3

And the probability of rolling a 3, 4 or 5  in a single roll  =   (1/6) + (1/6) + (1/6)  = 3/6  = 1/2

Let b be the unkown base

We have that

(4b + 4) ( 5b + 5)  = 3b^3 + 5b^2 + 0b + 6     simplify

20b^2 + 40b + 20  = 3b^3 + 5b^2 + 6

3b^3 - 15b^2  - 40b  - 14   = 0

Solving this using the Rational Zeroes Theorem shows that the integer solution  for b  = 7

Proof

44₇  * 55₇   = 

(4 * 7 + 4) ( 5 * 7 + 5)  =  

32 * 40  =  1280₁₀

And

3506₇  =   3*(7)^3  + 5*(7)^2 + 0*7 + 6  = 1280₁₀

Also  converting 1280 from base 10 to base 7

1280  =  182 * 7  +  R 6

182  = 26 * 7  + R 0 

26  =  3 * 7  +  R 5

3  = 0 * 7  + 3

Reading the  remainders from bottom to top we have 3506

Call a, b, c the sides

a + b + c  = 12  

And by the triangle inequality....

a + b > c

a + c > b

b + c > a

The greatest side can only  = 5   and the least side has to  be > 1

a    b     c

5    5     2

5    4     3

4    4     4

And these are the only possibilities

Interoir angle of a regular polygon  =  [ n - 2 ] *180  / n     where n is the number of sides

Octagon  =  [8 - 2 ] * 180 / 8  =     3/4  * 180  =  135°

Hexagon =   [ 6 - 2 ]* 180 / 6  =  (2/3) * 180  = 120 °

So....the interior angle of a regular octagon exceeeds that of a regular hexagon by  15°

This compound inequality is solved in a similar fashion as normal inequalities:

The solutions, if you are wondering, is any value for p that satisfies its current restraints of 2 < p < 5.