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March 29, 2024, 12:35:52 am

Author Topic: VCE Methods Question Thread!  (Read 4802391 times)  Share 

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TyranT

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Re: VCE Methods Question Thread!
« Reply #18015 on: July 16, 2019, 05:18:32 pm »
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Hello, I have a question:
It says 'write the following expression as a single power: 7 root 7' To clarify that means 7 multiplied by the root of 7.

Any help would be much appreciated.
Thanks
I have an attachment of the question and another.

^^^111^^^

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Re: VCE Methods Question Thread!
« Reply #18016 on: July 16, 2019, 05:26:44 pm »
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Hello, I have a question:
It says 'write the following expression as a single power: 7 root 7' To clarify that means 7 multiplied by the root of 7.

Any help would be much appreciated.
Thanks
I have an attachment of the question and another.
Hi, Root of 7 can be written as 7 to the power of 1/2 ( using index laws the nth root of a number is simply the number to the power of 1/n). 7 can be written as 7 to the power of 1. Using index laws (a different one this time) we can simplify this. 7^1 * 7^(1/2) = 7^(3/2)  (use the index law that states that a^n * a^m = a^(n+m)). Hope that helped :)

Edit: Sorry did u need help with question d) as well?
« Last Edit: July 16, 2019, 05:28:55 pm by ^^^111^^^ »

TyranT

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Re: VCE Methods Question Thread!
« Reply #18017 on: July 16, 2019, 05:28:41 pm »
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Thank you

^^^111^^^

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Re: VCE Methods Question Thread!
« Reply #18018 on: July 16, 2019, 05:29:38 pm »
+1

pugs

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Re: VCE Methods Question Thread!
« Reply #18019 on: July 16, 2019, 05:36:07 pm »
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As the integral of f(x) = a, the integral of f(x/5) would be 5a I think.
This would mean the answer would be 2(5a + 15a) = 40a.
thanks for the response! ;D may i ask how does f(x/5) = 5a? because i thought that if you pull the 1/5 out from the f(x/5), the value would be multiplied by 1/5


2019 vce journal here

TyranT

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Re: VCE Methods Question Thread!
« Reply #18020 on: July 16, 2019, 05:54:20 pm »
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I have another question i need help with. IDK why i am getting a different answer to the textbook.  Photomath is giving a different answer to the textbook and me.

Thanks.

MB_

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Re: VCE Methods Question Thread!
« Reply #18021 on: July 16, 2019, 06:21:16 pm »
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I have another question i need help with. IDK why i am getting a different answer to the textbook.  Photomath is giving a different answer to the textbook and me.

Thanks.
What answer are you getting?
2015-16: VCE
2017-: BSci UoM - Maths & Psych

^^^111^^^

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Re: VCE Methods Question Thread!
« Reply #18022 on: July 16, 2019, 06:32:34 pm »
+1
I have another question i need help with. IDK why i am getting a different answer to the textbook.  Photomath is giving a different answer to the textbook and me.

Thanks.
The answer should be A.  See my explanation below:
Let's first simplify what is inside the brackets of the first fraction. (using index laws b^3 will become b^1, as a^m / a^n = a^(m-n)). So then it will be (a^1/2 * b^2)^2. Using index laws again, [a^m * a^n = a^(m+n)] we can simplify this to a^1(or just a) * b^2. As division is involved, the second fraction will be flipped and will become b^4 divided by b^(3/2)*a. Now just multiply the numerators and it will become [a*(b^6)]! divided by [b^(3/2) * a]. The a will be cancelled (as there is a in the numerator and the denominator)@. This leaves us with b^6 in the numerator and b^3/2 in the denominator. Using division index law we can simplify this to b^(9/2).

! means use multiplication index law to do this
@ means use division index law to do this
[sorry I really should learn latex to actually do this in mathematical language - as I like to call it]. Honestly, please tell me if u don't understand.

TyranT

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Re: VCE Methods Question Thread!
« Reply #18023 on: July 17, 2019, 08:38:01 am »
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I was getting 9 but 111 helped me out

pugs

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Re: VCE Methods Question Thread!
« Reply #18024 on: July 18, 2019, 09:27:31 pm »
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if anyone would be able to help me out with this probability question, that'd be amazing!

A die is weighted as follows:
Pr(2) = Pr(3) = Pr(4) = Pr(5) = 0.2,    Pr(1) = Pr(6) = 0.1
The die is rolled twice, and the smaller of the numbers showing is noted. let Y represent this value,

Find Pr(Y = 1)

thank you in advance!


2019 vce journal here

redpanda83

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Re: VCE Methods Question Thread!
« Reply #18025 on: July 18, 2019, 09:40:25 pm »
+1
thanks for the response! ;D may i ask how does f(x/5) = 5a? because i thought that if you pull the 1/5 out from the f(x/5), the value would be multiplied by 1/5
Easiest way to explain -
f(x/5) - f(x) was dialated by factor of 5 parallel to x axis/from y axis.
      => means a became 5a after dialation. and area was dialated by factor of 5 as well . So integral of f(x/5) = 5a
         
VCE 2017 - 2019  |  ATAR: 97.65
Further [43], Chemistry [42],Physics[41], Methods [41],Hindi[35], EAL [39]

UoM 2020 - 2025  |  Bachelor of Science / Master of Engineering

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #18026 on: July 18, 2019, 09:42:49 pm »
+1
if anyone would be able to help me out with this probability question, that'd be amazing!

A die is weighted as follows:
Pr(2) = Pr(3) = Pr(4) = Pr(5) = 0.2,    Pr(1) = Pr(6) = 0.1
The die is rolled twice, and the smaller of the numbers showing is noted. let Y represent this value,

Find Pr(Y = 1)

thank you in advance!

First, let's write all the possible outcomes that make up the event \(Y=1\). They are: \[(1,1),\ (1,2),\ (2,1),\ (1,3),\ (3,1),\ (1,4),\ (4,1),\ (1,5),\ (5,1),\ (1,6),\ (6,1).\] We can write down the probability of each outcome and add them up to obtain the required result. Note that some of the outcomes have the same probability \((\)eg:  \(\Pr(1,2)=\Pr(2,1)\)\(\,)\)
\[\Pr(Y=1)=(0.1\times 0.1)+2(0.1\times 0.2)+2(0.1\times 0.2)+2(0.1\times 0.2)+2(0.1\times 0.2)+2(0.1\times 0.1)=\boxed{0.19\,}\]
2015\(-\)2017:  VCE
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pugs

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Re: VCE Methods Question Thread!
« Reply #18027 on: July 19, 2019, 06:57:50 pm »
+1
thank you so much redpanda83 and AlphaZero!!!


2019 vce journal here

persistent_insomniac

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Re: VCE Methods Question Thread!
« Reply #18028 on: July 21, 2019, 06:44:42 pm »
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Hey  can someone please help me with this question:
dQ/dt = 9e^-0.1t. Show that dQ/dt = (100-Q)/10.

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #18029 on: July 21, 2019, 10:49:23 pm »
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Hey  can someone please help me with this question:
dQ/dt = 9e^-0.1t. Show that dQ/dt = (100-Q)/10.

Is there more to this question? On its own, you do not have enough to say that \[\frac{dQ}{dt}=9e^{-t/10}\implies \frac{dQ}{dt}=\frac{100-Q}{10}.\] From the information given, we have \[Q=\int 9e^{-t/10}=-90e^{-t/10}+C,\quad C\in\mathbb{R},\]and so if it is the case that \(\dfrac{dQ}{dt}=\dfrac{100-Q}{10}\), we would require \(C=100\) (check this yourself), but we don't have this information.
2015\(-\)2017:  VCE
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