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#### fun_jirachi

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##### Re: General Math Q+A Thread
« Reply #645 on: October 24, 2018, 07:42:58 pm »
0
Hey there!

For the first one, notice that all the units in the question and answers are in SI units! It's not specifically maths, but 1 watt (W) is equal to 1 Joule per second (Js-1), so the answer is C.

For the second and third one, I've both downloaded and expanded the pictures and I still can't read them! Do you happen to have a clearer picture, because if so I'd be glad to help you.

For the last one, its just knowing that tera- means roughly 1012. Knowing that there are 1024 bytes in a megabyte and 1024 megabytes in a gigabyte and so on as well as that 8 bits make one byte, 23(for the bits)x10244 (since one terabyte is roughly 1012, or think about it as 1024 and going up 4 levels). 1024 = 210 so using indice laws you get that there are 243 bits in 2 terabytes.

Hope this helps!

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#### jamonwindeyer

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##### Re: Standard Math Q+A Thread
« Reply #646 on: November 10, 2018, 04:25:23 pm »
+5
Hey! With HSC exams now over, we are now doing the new version of General Maths - Standard Maths! Almost everything about the old course has stayed the same, so everything from before this point is still super useful. Feel free to have a scroll back!
« Last Edit: November 10, 2018, 04:27:01 pm by jamonwindeyer »

#### Equilibrium

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##### Re: Standard Math Q+A Thread
« Reply #647 on: November 18, 2018, 08:17:36 pm »
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Hi! This is my second post on here, I'm in Year 10 & about to transition into Year 11. I've already elected to undertake General Maths because I know how important it is to learn... But I'm currently reading a textbook for next year & cannot comprehend what I'm reading. I've never been good at maths, always easily getting better marks in my language & humanities based subjects.
I'm too embarrassed to ask a teacher about it, I don't want someone I might face every day thinking I'm dumb...
Would anyone be able to help me decipher this? 😅

“The inverse operation to exponentiation is logarithms, and as exponents aren't commutative like addition and multiplication are, there are two possible inverses, the other being surds.
The square root of (x squared) equals x, and so in that way the root is the inverse operation, but then there's also this (the first equation has been shifted up by 1, but both essentially give the graph for y=x, despite the second equation including an exponent).”

There's a graph diagram underneath.

#### S200

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##### Re: Standard Math Q+A Thread
« Reply #648 on: November 18, 2018, 08:40:15 pm »
+3
I don't fully get what it's saying myself (maybe if you uploaded a picture of the graphs we could help better?) but I'll try to explain a little.
“The inverse operation to exponentiation is logarithms, and as exponents aren't commutative like addition and multiplication are, there are two possible inverses, the other being surds.
$y=2^x$ is an exponential, and you would get the inverse by placing the $y$ where the $x$ is.
Solving for $y$ requires using a logarithm.
The Basic form of a logarithm
$y=a^x \therefore \log_a {y}=x$
Eg: $64=2^x \quad \log_2 {64} = x, \quad x=8$
So, when solving the aforementioned equation, we would find that the inverse function to $y=2^x$ is $y=\log_2 {x}$.

We can visually see that this is correct when we graph it, by inserting the line $y=x$, showing that the reflection is symmetrical about this line.

By commutative, the textbook is saying that unlike addition and multiplication, where $1+2=2+1$ and $a\times b=b\times a$, an exponential cannot be the so easily switched around...

i.e: $2^x \ne x^2$

Please reply if anything doesn't make sense, and I'll try to help more.
Also, Welcome to AN...
« Last Edit: November 18, 2018, 08:43:39 pm by S200 »
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#### ChloeWalker

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##### Re: Standard Math Q+A Thread
« Reply #649 on: January 15, 2019, 01:43:51 pm »
+1
Hi, I'm not too sure if this is where I can ask this, if it's not if someone could redirect me that would be great!

It's a standard maths question:

"Calculate the future value, to the nearest cent.  a) Present value of $680 invested for 4 years at 5% p.a. compounded biannually" the answer in the book is$828.51 (Cambridge maths book)

This is what I did and if someone can spot the flaw please let me know!!
P = 680
r = 10%
period = 2

Thankkss!!!!

#### RuiAce

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##### Re: Standard Math Q+A Thread
« Reply #650 on: January 15, 2019, 01:56:28 pm »
+3
Hi, I'm not too sure if this is where I can ask this, if it's not if someone could redirect me that would be great!

It's a standard maths question:

"Calculate the future value, to the nearest cent.  a) Present value of $680 invested for 4 years at 5% p.a. compounded biannually" the answer in the book is$828.51 (Cambridge maths book)

This is what I did and if someone can spot the flaw please let me know!!
P = 680
r = 10%
period = 2

Thankkss!!!!
It's a trap in the wording.

For varying reasons that I never quite understood myself, "biannually" actually means compounded twice every year, as opposed to once every two years. Which is equivalent to saying that the interest is actually compounded semi-annually. So you should have $r = 2.5\%$ and $\text{number of periods} = 8$, giving $680(1.025)^8$ as your answer.

#### alole201

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##### Re: Standard Math Q+A Thread
« Reply #651 on: February 08, 2019, 10:30:41 pm »
+1
Hi everyone!!

I was wondering if anyone could help me with explaining bearings (true, compass etc.). I've been confused for the past few days about it and i was hoping for some clarification!

Thanks

#### ChloeWalker

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##### Re: Standard Math Q+A Thread
« Reply #652 on: March 03, 2019, 08:08:03 pm »
0
Hi can anyone help me out with this question:

"How long will it take to pay off a credit card debt of \$1000 if the annual interest rate is 18.5% and you only make the minimum payment each month (assuming no other transactions)? How much interest will you pay?"

#### matthewwu2003

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##### Re: Standard Math Q+A Thread
« Reply #653 on: July 10, 2019, 04:25:30 pm »
+1
Hi everyone!!

I was wondering if anyone could help me with explaining bearings (true, compass etc.). I've been confused for the past few days about it and i was hoping for some clarification!

Thanks

There are two types of bearings, True Bearing and Compass Bearing.

Compass Bearing specifies the direction in terms of the four cardinal directions: North (N), South (S), East (E), West (W). All other directions are indicated by the deviation from North (N) or South (S) towards the East (E) or West (W).
For Example: N35E, S41W, N23W

True Bearing specifies the direction by its deviation from North. It is measured clockwise from North and expressed as a three digit number.
For Example: 035T, 056T, 120T

Below I have attached some diagrams for your reference.

#### LoneWolf

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##### Re: Standard Math Q+A Thread
« Reply #654 on: July 18, 2019, 06:01:11 pm »
0
Hey I know this is standard and I do mathematics (year 11) but i have forgotten the null factor law. Can someone explain how i solve dydx=0 to find x & y coordiates in view of finding where any points where the tangent is horizontal:

y= (x^2-1)^3
dy/dx = 6x(x^2-1)^2
dy/dx = 0
....
....
....

& i am confused from here!
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#### fun_jirachi

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##### Re: Standard Math Q+A Thread
« Reply #655 on: July 18, 2019, 08:17:48 pm »
+2
Hey there!

Totally okay to post in the 2U thread btw

Basically the null factor law states that if at least one of the factors (in this case of the given polynomial) of some real number is zero, then the product is zero. Conversely, if all the factors are nonzero, then the product is nonzero.

Take for example the following polynomial
$f(x)=(x-a_1)(x-a_2)(x-a_3)...(x-a_n) \\ \text{(Where} \ a_1, a_2, a_3 ... a_n \ \text{are distinct and real)} \\ \text{If we take the function of any one of} \ a_1, a_2, a_3 ... a_n \ \text{we have a root} \\ ie. f(x)=0$

$6x(x^2-1)^2=0 \\ 6x[(x+1)(x-1)]^2=0 \\ \text{We find that} \ f'(x)=0 \ \text{when} \ x=0, \pm 1$
ie. there are horizontal tangents at (0, -1), (1, 0) and (-1, 0)

Hope this helps
« Last Edit: July 19, 2019, 11:31:28 am by fun_jirachi »
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#### LoneWolf

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##### Re: Standard Math Q+A Thread
« Reply #656 on: July 19, 2019, 09:25:05 am »
+1
Ok sweet.
Just 1 thing did you mean (-1,0) rather than (-1,-8)?
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#### fun_jirachi

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##### Re: Standard Math Q+A Thread
« Reply #657 on: July 19, 2019, 11:32:07 am »
0
Ok sweet.
Just 1 thing did you mean (-1,0) rather than (-1,-8)?

Yep, thanks for the pick-up Post has been edited to reflect that
Failing everything, but I'm still Flareon up.

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#### alole201

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##### Re: Standard Math Q+A Thread
« Reply #658 on: July 25, 2019, 05:29:13 pm »
0

An experiment has three distinct outcomes, A, B and C. Outcome A occurs 50% of the
time. Outcome B occurs 23% of the time.

What is the expected number of times outcome C would occur if the experiment is
conducted 500 times?

A. 115
B. 135
C. 250
D. 365
thank you!

#### fun_jirachi

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##### Re: Standard Math Q+A Thread
« Reply #659 on: July 25, 2019, 06:04:54 pm »
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Hey there!

If n distinct outcomes are possible for some experiment, it's implied that the sum of the probabilities a1+a2+a3+...+an = 1, where ak is one of the distinct outcomes' probabilities, where k is an integer such that 0<k<=n.

As such, in this question, we have three distinct outcomes whose probabilities must sum to 1, or 100%. Given the other two probabilities, we can subtract them from 100%, and thus we obtain the fact that C occurs 27% of the time. Hence, if the experiment is conducted 500 times, outcome C will occur in 27% of 500 times, ie. the answer should be B, 135.

Hope this helps
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Advanced [87] | Maths Extension 1 [98] | Maths Extension 2 [97]
ATAR: 99.05

UCAT: 3310 - Verbal Reasoning [740] | Decision Making [890] | Quantitative Reasoning [880] | Abstract Reasoning [800]