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Obviously, the growth is an exponential growth, as the modelling function suggests.

A "strange" thing to say is that, year 2000 is 0 years after year 2000.

So we substitute t = 0, and we get:

Population living in the city in 2000 = \(2e^{0.06(0)} = 2(1) =2\text{ millions}\)

\(\text{Consider the range of }\tan^{-1} x.\\ \tan^{-1} x \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]\\ \text{That means we need to find a value } v\text{, which is between}-\dfrac{\pi}{2}\text{ and }\dfrac{\pi}{2}\text{, such that}\tan v = \tan\left(\dfrac{4\pi}{5}\right)\\ \)

\(\text{Solving the equation, }\\ v = \dfrac{4\pi}{5} + n\pi, \; n\in \mathbb Z\)

Only when n = -1, v is inside the required range.

The solution we need is \(v = \dfrac{4\pi}{5} - \pi = \dfrac{-\pi}{5}\).

Therefore \(\tan^{-1}\left(\tan\left(\dfrac{4\pi}{5}\right)\right) = \dfrac{-\pi}{5}\)

4(c)

\(\text{As shown above in 4(b), } \cos\left(2x-\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{3}\right)\implies x = n\pi+\dfrac{\pi}{2}\text{ or }x=n\pi+\dfrac{\pi}{3}\\ \text{As tan} x \text{ has a period of }\pi,\\ \tan x = \tan\left(\dfrac{\pi}{2}\right)\text{ or }\tan x = \tan\left(\dfrac{\pi}{3}\right)\\ \text{But considering the fact that}\tan x \text{ has an asymptote at } x = \dfrac{\pi}{2},\text{ the only finite value is }\tan\left(\dfrac{\pi}{3}\right).\\ \text{The required value is }\tan\left(\dfrac{\pi}{3}\right) = \sqrt3\)

4(b)

\(\cos\left(2x-\dfrac{5\pi}{6}\right) = \sin\left(\dfrac{\pi}{3}\right)\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}\right)\\ \text{As the cosine function is periodic with period }2\pi\text{, }\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(\dfrac{\pi}{6}+2n\pi\right)\\ \text{As }\cos x = \cos\left(-x\right)\text{, }\\ \cos\left(2x-\dfrac{5\pi}{6}\right) = \cos\left(-\dfrac{\pi}{6}+2n\pi\right)\\ \text{Comparing the left hand side and the right hand side of the equation, }\\ 2x-\dfrac{5\pi}{6} = \dfrac{\pi}{6} + 2n\pi \text{ or } 2x-\dfrac{5\pi}{6} = -\dfrac{\pi}{6} + 2n\pi\\ 2x = \pi+2n\pi\text{ or }2x=\dfrac{2\pi}{3}+2n\pi\\ x = \dfrac{\pi}{2}+n\pi\text{ or }x=\dfrac{\pi}{3}+n\pi\)

What does $ mean?

This is trigonometry.

4(a)

\(\text{As }\sin x = \cos\left(\dfrac{\pi}{2}-x\right),\\ \sin \left(\dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{2}-\dfrac{\pi}3\right)=\cos\left(\pi\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\right)=\cos\left(\pi\left(\dfrac{1}{6}\right)\right) =\cos\left(\dfrac{\pi}{6}\right)\\ \sin \left(\dfrac{\pi}{3}\right) = \cos\left(\dfrac{\pi}{k}\right)\\ \cos\left(\dfrac{\pi}{6}\right) = \cos\left(\dfrac{\pi}{k}\right)\\ \text{Comparing the left hand side and the right hand side of the equation,}\\ k =6\)

Please notice that \((a+b)^n \neq a^n+b^n\).

It is easy enough, even with that incorrect formula :)

The correct formula should be \(\displaystyle \sum^{n}_{i=1} i^2 = \dfrac{n(n+1)(2n+1)}{6}\).

Obviously, this limit is a Riemann integral.

\( S = \displaystyle{\lim_{n\rightarrow \infty} \sum_{i=1}^{n}\left(\left(\dfrac{2}{n}\right)\left(\dfrac{i}{n}\right)^2\right)\\ \;\;\;\!= 2\lim_{n\rightarrow \infty} \sum_{k=1}^{n}\left(\left(\dfrac{1}{n}\right)\left(\dfrac{k}{n}\right)^2\right)\\ \boxed{\text{Riemann integral: For any function f continuous on [0,1], } \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\dfrac{1}{n}f\left(\dfrac{k}{n}\right) = \int^1_0f(x)dx }\\ S=2\int^1_0x^ 2dx\\\;\;\;\!=2\left(\dfrac{1}{3}\right)\\\;\;\;\!=\dfrac{2}{3} }\)

\(V = \dfrac{1}{3}Ah\)

\(A = 32 \cdot 66 = 2112\)

\(h = \sqrt{65^2-\left(\dfrac{32}{2}\right)^2-\left(\dfrac{66}{2}\right)^2} = 24\sqrt5\)

\(V = \dfrac{1}{3}(2112)(24\sqrt5) = 16896\sqrt5\)

Thank you for your help!

\(f(x) = q(x)d(x)+r(x)\)

deg f = 9, deg r = 3, deg d > deg r.

For maximum deg q. deg d is minimum. => deg d = 4.

\(O(x^9) = q(x) O(x^4) + O(x^3)\)

q(x) = O(x⁵)

Therefore deg q = 5.

\(P(x) = 3x^3+a_2x^2+a_1x-6\)

Product of roots = \(-\dfrac{-6}{3}\)= 2.

Possible roots = {1,-1,2,-2}.

\(\quad (\sqrt[3]{7}+\sqrt[3]{49})^3\\ = 7 + 49 + 3\sqrt[3]{7\cdot49}(\sqrt[3]{7}+\sqrt[3]{49})\\ = 56 + 21(\sqrt[3]{7}+\sqrt[3]{49})\)

Therefore P(x) is x³ - 21x - 56.

Product of roots = \(-\dfrac{-56}{1} = 56\)

The tan inverse and the tangent just 'cancel' each other and you are left with  4pi/5....

   DO you see?   First you do the tangent.....then you do the tangent inverse...

        here is another example    sin^-1 ( sin 30)   =    sin^-1 ( 1/2 ) = 30

L = [ (150^(2/3) + 200^(2/3) ] ^(3/2)    = (28.231  + 34.1996)^(3/2) =   62.43^(3/2) = 493.28 cm

Thanks alot'

The amount of algae covering the Smith's backyard pond doubled every day until it was completely covered in algae on day 30 of the month. On what day of that month was 75% of the pond algae-free?

On day 30 it was 100% covered.

On day 29 it was 50% covered.

On day 28 it was 25% covered, which is the same as 75% algae-free.

?

what do you need help with?

Good job, GM!!!!

P ( Burgers) =  89 / 153

1)

You could do this problem the rigorous way and look up the probability of the high and low z-scores,

but because each of the bounds are integer multiples of the standard deviation, -2, and 2, about the mean,

I suspect they want you to know that

\(P[(\mu -\sigma,\mu+\sigma)]\approx 0.68\\ P[(\mu -2\sigma,\mu+2\sigma)] \approx 0.95\\ P[(\mu-3\sigma, \mu+3\sigma)] \approx 0.997\\ \text{etc.}\)

So we are looking at 95% of the total population of 300 gamers.

This is 285 people.