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Chikis

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  1. 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?

  2.  

    Yet your response was to introduce a universal set, that was not correct, and not called for by that part of the question.

    Did I do wrong to introduce the universal set?

     

    But they are not sets of the same types of number.

    Universal set may contain diffrent number which are elements of different subsets.

     

    Q contains all the odd integers, excluding 0, which is neither odd nor even.

    Are you saying that zero is not an even numbers? I believe these are lists of even numbers : 0, 2, 4, 6, 8, 10, ...,

     

     

    Can you now state this in symbols?

    Let me try to be specific now. I can find

     

    [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.

  3. [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]

  4. 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]?

  5. 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

  6. 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.

  7. 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.

  8. My main aim of starting this thread has not been met.

    Is your final line intended to be an answer or just a point on way to answer?

    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.

    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?

  9. 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?

  10. So 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.

    Do you mean

    [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.

    So 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.

    Do you mean

    [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.

  11. 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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