Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie

Dreading ordinary level maths!!!!

124»

Comments

  • Registered Users, Registered Users 2 Posts: 9,034 ✭✭✭Ficheall


    how does one do differentiation?

    Some very basic qs attached - should hopefully cover you for the parts (a) though.


  • Registered Users, Registered Users 2 Posts: 9,034 ✭✭✭Ficheall


    marko93 wrote: »
    Attachment wont work D:

    Really?

    hmm

    x^\frac{a}{b}=\sqrt{x^a}$\\
    So $8^\frac{1}{3}=\sqrt[3]{8^1}$\\
    =\sqrt[3]{8}=2$\\
    Similarly, $4^\frac{1}{4}=\sqrt{2}$ (ie. $2^\frac{1}{2}$)\\
    Note that $x^a\times x^b=x^{a+b}$,
    so
    \[(8^\frac{1}{3})(4^\frac{1}{4})=2^{5-x}\]
    filling in the answers gotten from parts i and ii
    \[(2)(\sqrt{2})=2^{5-x}\]
    \[(2^{1.5}=2^{5-x}\]
    So $1.5=5-x$ giving $x=3.5$

    Any good to you?

    edit:
    x^(a/b)=bth root of (x^a)
    So 8^(1/3)=Cube root of(8^1)
    =Cube root of (8)=2
    Similarly, 4^(1/4)=sqrt(2) (ie. 2^(1/2))

    Note that (x^a)(x^b)=x^(a+b),
    so
    (8^(1/3))(4^(1/4))=2^(5-x)
    filling in the answers gotten from parts i and ii
    (2)(\sqrt(2))=2^(5-x)
    2^(1.5)=2^(5-x)
    So 1.5=5-x giving x=3.5


  • Registered Users, Registered Users 2 Posts: 4,010 ✭✭✭marko93


    Kinda D:
    All the different signs are confusing though


  • Registered Users, Registered Users 2 Posts: 9,034 ✭✭✭Ficheall


    I tried to englishify it a little in the edit... it looks much nicer latexed..


  • Registered Users, Registered Users 2 Posts: 4,010 ✭✭✭marko93


    Ficheall wrote: »
    Really?

    hmm

    x^\frac{a}{b}=\sqrt{x^a}$\\
    So $8^\frac{1}{3}=\sqrt[3]{8^1}$\\
    =\sqrt[3]{8}=2$\\
    Similarly, $4^\frac{1}{4}=\sqrt{2}$ (ie. $2^\frac{1}{2}$)\\
    Note that $x^a\times x^b=x^{a+b}$,
    so
    \[(8^\frac{1}{3})(4^\frac{1}{4})=2^{5-x}\]
    filling in the answers gotten from parts i and ii
    \[(2)(\sqrt{2})=2^{5-x}\]
    \[(2^{1.5}=2^{5-x}\]
    So $1.5=5-x$ giving $x=3.5$

    Any good to you?

    edit:
    x^(a/b)=bth root of (x^a)
    So 8^(1/3)=Cube root of(8^1)
    =Cube root of (8)=2
    Similarly, 4^(1/4)=sqrt(2) (ie. 2^(1/2))

    Note that (x^a)(x^b)=x^(a+b),
    so
    (8^(1/3))(4^(1/4))=2^(5-x)
    filling in the answers gotten from parts i and ii
    (2)(\sqrt(2))=2^(5-x)
    2^(1.5)=2^(5-x)
    So 1.5=5-x giving x=3.5
    Wait, wait, wait.... Now it makes more sense D:
    Thank you so much :D


  • Advertisement
  • Registered Users, Registered Users 2 Posts: 2,292 ✭✭✭LilMissCiara


    marko93 wrote: »
    Wait, wait, wait.... Now it makes more sense D:
    Thank you so much :D

    Go get your log tables right now and make sure you are familiar with what's where!

    Not wanting to offend anybody but I know that the students in one of the ordinary level classes in my school have never owned a log tables. The teacher photocopies what they need for each chapter and for the Summers and Christmas tests and stuff the teacher includes photocopies of the pages in the log tables that they need! I think it's mad!

    So yeah, look at your log tables!


  • Registered Users, Registered Users 2 Posts: 4,010 ✭✭✭marko93


    In relation to that question! I DONT EVEN KNOW WHERE I SHOULD LOOK D:
    My teacher never went throught the Log tables with us:/


  • Registered Users, Registered Users 2 Posts: 2,292 ✭✭✭LilMissCiara


    marko93 wrote: »
    In relation to that question! I DONT EVEN KNOW WHERE I SHOULD LOOK D:
    My teacher never went throught the Log tables with us:/

    Haven't mine with me but there is an index at the start. Maybe go to school a bit early and try and borrow a set of somebody just to have a look at them.

    Everything for maths is at the start however there is some tax formulas which I think you use in ordinary level. They are under the economics section!


  • Registered Users, Registered Users 2 Posts: 4,010 ✭✭✭marko93


    Haven't mine with me but there is an index at the start. Maybe go to school a bit early and try and borrow a set of somebody just to have a look at them.

    Everything for maths is at the start however there is some tax formulas which I think you use in ordinary level. They are under the economics section!
    I have my own! Just barely use them :L
    I'll have a glance through though :D


  • Registered Users Posts: 348 ✭✭xclw


    how'd paper 1 go for everyone?


  • Advertisement
  • Moderators, Category Moderators, Education Moderators Posts: 27,232 CMod ✭✭✭✭spurious


    AdamD wrote: »
    Not to sound arrogant, but how does one fail ordinary level maths? :eek:

    Major fail on the first part of your post.

    Oh how lovely it must be in your high kingdom, where Maths is easy.

    If you haven't anything useful to say, stay quiet.


This discussion has been closed.
Advertisement