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Posts posted by Chikis
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Please I need help with these problem:
A rectangular courtyard 8m by 6m has a rectangular lawn of area 12m^2 situated in it in such way as to leave a path x m wide all round the inside of the courtyard. Find the value of x.
From the way, I represented the xm wideness all around inside the courtyard, it shows that the area the lawn=(8-2x)m by (6-2x)m = 12m^2
The area of only the lawn = 12m^2
The area of the courtyard = 48
48-12 = 36
The 36m^2 is what area? Is it area of the path?
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Did I do wrong to introduce the universal set?Yet your response was to introduce a universal set, that was not correct, and not called for by that part of the question.
Universal set may contain diffrent number which are elements of different subsets.But they are not sets of the same types of number.
Are you saying that zero is not an even numbers? I believe these are lists of even numbers : 0, 2, 4, 6, 8, 10, ...,Q contains all the odd integers, excluding 0, which is neither odd nor even.
Let me try to be specific now. I can findCan you now state this in symbols?
[math](b)P\cap{R}[/math]
[math](c )(P\cap{Q}[/math] without introducing a universal set. My main problem is how to find
[math](b)P'\cap{R'}[/math]
[math](c )(P\cap{Q)'}[/math]
I had that problem because of the compliment involved and there is no universal set introduced. So I had to introduce one if I must tackle the problem.
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[math]U=\{...,-2,-1,0,1,2,...,7,9,11,13,....\}[/math]
Am thinking that all the subsets are contained in the universal set. I can hardly figure what is wrong what in the universal set. Please help me. I have really suffered on this.
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What is wrong here? The whole thing seems right to me. Could you help me figure out what is wrong?You need to be consistent
One of the above is clearly wrong.
Have a read here - http://www.mathsisfun.com/sets/venn-diagrams.html
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[math]P=\{...,-3,-2,-1,0\}[/math][math]Q=\{...,-5,-3,-1,1,3,5,...\}[/math][math]R=\{-2,-1,0,1,2,...,6\}[/math]based on the above subsets[math]U=\{...,-2,-1,0,1,2,...,7,9,11,13,....\}[/math]
[math]\therefore[/math][math]Q\cap R=\{-1,1,3,5\}[/math]
[math]P'=\{1,2,...6,7,9,11,...\}[/math]
[math]R'=\{...,-5,-4,-3,7,9,11,13,...\}[/math][math]P'\cap R'=\{7,9,11,13,...\}[/math]
[math]P\cap Q=\{...,-7,-5,-3,-1\}[/math][math](P \cap Q)'=\{...,-8,-6,-4,-2,0,1,2,3,...,7,9,11,...\}[/math]
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Must I draw a the diagram to solve this problem? I feel that is not necessary. I just need to understand this problem.
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Fourthly, try to understand what P' n R' and (PnQ)' actually mean - ie everything which is outside P and at the same time outside R, everything which is not in both P and Q.
So say for example does P' mean opposite or reverse of all the element in set P?
Let me see if I can interprete what
[math]P=\{x : \le0\}[/math]and[math]R=\{x: -2\le x<7\}[/math] mean respectively.
Could [math]P=\{x : \le0\}[/math] mean
[math]P=\{....-3,-2,-1,0\} [/math]
and [math]R=\{x: -2\le x<7\}[/math] mean[math]\{-2,-1,0,1,2,3,4,5,6\}[/math]?
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My first problem concerning this problem:
Do these mean the same thing?
1[math]P'\cap R'[/math]and[math](P\cap R)'[/math]
2[math]( P\cap Q)'[/math]and[math]P'\cap Q'[/math]
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Never mind. This [math]( c )(P \cap Q)'[/math] has sufficed everything.
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Not untill when it becomes[math]( C )(P \cap Q)[/math]Something like the above?
[math]( c )(P \cap Q)'[/math]
that c is a small letter. Is just like a kind of numbering the problems. We have problems a, b and c.
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Solution A is 0.10moldm-3 HCl. B is a solution of ... Solution A is 0.10moldm-3 HCl. B is a solution of sodium trioxocarbonate(IV). 25cm3 of B is titrated against A using methyl orange as indicator. 23.5cm3 of A is used in titration. Calculate the mass of sodium trioxocabonate(IV) present in one dm3 of the solution B. [Na = 23, C = 12, O = 16]
molar concentration of acid, CA =[math]0.10moldm^{-3}[/math]
volume of acid, VA = [math]23.5cm^3=0.0235dm^3[/math]
molar concentration of base, CB=?
volume of base, VB = [math]25cm^3=0.025dm^3[/math]
from equation of reaction:
[math]HCl(aq)+Na_2CO_3(aq)\to2NaCl(aq)+H_2O(l)+CO_2(g)[/math]
we can see that
a/b = mole ratio of acid to base
[math]\therefore[/math] a:b = 2:1
[math]\to[/math]
Solution A is 0.10moldm-3 HCl. B is a solution of ... Solution A is 0.10moldm-3 HCl. B is a solution of sodium trioxocarbonate(IV). 25cm3 of B is titrated against A using methyl orange as indicator. 23.5cm3 of A is used in titration. Calculate the mass of sodium trioxocabonate(IV) present in one dm3 of the solution B. [Na = 23, C = 12, O = 16]
molar concentration of acid, CA =[math]0.10moldm^{-3}[/math]
volume of acid, VA = [math]23.5cm^3=0.0235dm^3[/math]
molar concentration of base, CB=?
volume of base, VB = [math]25cm^3=0.025dm^3[/math]
from equation of reaction:
[math]HCl(aq)+Na_2CO_3(aq)\to2NaCl(aq)+H_2O(l)+CO_2(g)[/math]
we can see that
a/b = mole ratio of acid to base
[math]\therefore[/math] a:b = 2:1
[math]\to[/math]
[math]CB=\frac{bCAVA}{aVB}=\frac{0.10\times0.0235}{2\times0.025}=0.047moldm^{-3}[/math]
molar mass of [math]Na_2CO_3=(23\times2)+12(16\times3)=106gmol^{-1}[/math]
The problem now is how do I get the mass of sodium trioxocarbonate(IV) present in one [math]dm^3[/math] of the solution B?
Moderator please help me remove the other two redundant threads
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Yes. You got it.Chikis - I fixed your latex. Hope I got the expression right!
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Mind you, this is what I actually want to write.[math]\vert A \vert[/math], [math]\textnormal{n}(A)[/math], [math]\textnormal{card}(A)[/math], [math]\# A[/math], [math]\bar{\bar{A}}[/math].
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Men! I have a big problem here oh....
I find it difficult rendering number of elements in a set. For example:[math]n(A)[/math]
But that's not what I want to write. I want to have this shown in latex: ©(P\cap{Q})'
I wrote it like this:[math](c)(P\cap{Q})'[/math] and am having latex error. I have gone through the whole library of latex symbols;no help is sight. I need help please.
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The universal set is the set of all interger and the subsets P,Q,R of are given by
[math]P=\{x : \le0\}[/math]
[math]Q=\{....,-5,-3,-1, 1, 3, 5....\}[/math]
[math]R=\{x: -2\le x<7\}[/math]
Find:
[math](a)Q\cap{R}[/math]
[math](b)P'\cap{R'}[/math]
[math](c )(P\cap{Q)'}[/math]
I need guidance in tackling this problem.
Moderator please help me remove the other two redundant threads.0 -
The book says the answer is 9.46.
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My main aim of starting this thread has not been met.
I feel that am not handling the question the way I should, so I need some body to help get me to the right part to get this problem solved.Is your final line intended to be an answer or just a point on way to answer?
I want to explore this problem more and see what will come out as the final solution. This is because the result that I get by the time I evaluate [math]10^{2.3675}+10^{0.9750}[/math] is very different from the answer key in my textbook. I want to understand what is going on. Is my approach wrong? Am I not doing it the right way? What is the problem?
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Thanks for voluntering to help. I have reasoned and taught about the problem in a more better way.
30% of y = x
[math]\frac{3}{10}\times{y}=x[/math]
[math]\to[/math]
3y = 10x
It is now easier to find 30% of 3y and see what it equals.
[math]\therefore[/math]
[math]3y\times\frac{3}{10}=10\times\frac{3x}{10}[/math]
So we can clearly now that
[math]3y\times\frac{3}{10}=3x[/math]
I want to use this opportunity to ask a question concerning the use of latex to render cancellation.
For example, I have [math]10\times\frac{3x}{10}[/math], how do use latex to show that 10 cancelled 10 so that 3x is left.
I mean I want to draw backlash or stroke each on both the 10 at the numerator and denominator respectively to show that. How do I do it? Which code should I use?
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Yes, oh! Am intrested in knowing or getting a better way to express it.I'm sure chikis would be interested in your better way, if you told us what that was.
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Do you meanSo you have an equation for x in terms of y
So rearrange it to get an equation for y in terms of x
Then multiply what ever this is by 3 to get 3y ie 3 times as much.
Then take 30% of this.
You should notice something interesting about the result.
[math]\frac{3}{10}\times{x}=x[/math]
[math]\to[/math]
y = 10x/3
multiplyiny by 3, we have 3y = 10x.
But why are we multiplying by 3?
Taking 30% of both sides, we have
[math]3y\times\frac{30}{100}=10x\times\frac{30}{100}[/math]
[math]\to[/math]
[math]3y\times\frac{30}{100}=\cancel{10}x\times\frac{3}{\cancel{10}}[/math]
This gives 30% of 3y in terms of x as 3x.
Do you meanSo you have an equation for x in terms of y
So rearrange it to get an equation for y in terms of x
Then multiply what ever this is by 3 to get 3y ie 3 times as much.
Then take 30% of this.
You should notice something interesting about the result.
[math]\frac{3}{10}\times{x}=y[/math]
[math]\to[/math]
y = 10x/3
multiplyiny by 3, we have 3y = 10x.
But why are we multiplying by 3?
Taking 30% of both sides, we have
[math]3y\times\frac{30}{100}=10x\times\frac{30}{100}[/math]
[math]\to[/math]
[math]3y\times\frac{30}{100}=10x\times\frac{3}{10}[/math]
This gives 30% of 3y in terms of x as 3x.
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So what do I do next?When no base is given, the convention is to assume 10.
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Sorry, I intended writingThat is not an equation.
[math]\frac{3}{10}\times{y}=x[/math]
Am saying in the equation above, what is in terms of what? And why or how we do we know?
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If log x = 2.3675 and log y = 0.9750, what is the value of x+y correct to three significant figures?
I take it that
[math]log_x=2.3675[/math] gives
[math]x=10^{2.3675}[/math]
and that
[math]log_y=0.9750[/math]
[math]\to[/math]
[math]y=10^{0.9750}[/math]
x+y gives [math]10^{2.3675}+10^{0.9750}[/math]
Is my thinking true?
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How do I show longrightarrow?
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Find the value of x.
in Homework Help
Posted · Edited by Chikis
I don't even have problem finding x. I want to verify something. If the area of path is 36m^2. What is the dimension of the path in terms of x?