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Math Help - Math B Regents Help?

  1. #1
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    Math B Regents Help?

    Hi, I was wondering if i can get some help for a few math B regents help. one problem that i dont get is from the june 2005 regents number 3
    question:http://www.nysedregents.org/testing/mathre/b605.pdf
    answer: http://www.nysedregents.org/testing/mathre/b-key605.pdf
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    Quote Originally Posted by shanegoeswapow View Post
    Hi, I was wondering if i can get some help for a few math B regents help. one problem that i dont get is from the june 2005 regents number 3
    question:http://www.nysedregents.org/testing/mathre/b605.pdf
    answer: http://www.nysedregents.org/testing/mathre/b-key605.pdf
    For this question, you are expected to know two things:

    (1) \sin x = \sin (180 - x)

    (2) \sin (-x) = - \sin(x)



    Thus \sin (-230) = - \sin(230) = - \sin (180 - 230) = - \sin (-50) = \sin (50)

    which is choice 1
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    thanks alot, that really helped
    another question from the same regents is number 10, im not so good at logs either
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    Quote Originally Posted by shanegoeswapow View Post
    thanks alot, that really helped
    another question from the same regents is number 10, im not so good at logs either
    Logs are exponents. there's nothing to be freaked out about.

    There are four laws you NEED to know in order to do pretty much any log problem. There are about 10 laws you need to know to do any log problem comfortably, that is, the other laws are derived from the first four, so knowing them gives you a shortcut.

    Here are the four laws you MUST know. the first law is the definition of a logarithm.


    LAW 1: The logarithm of a number, to a given base, is the power to which the base must be raised to give the number.

    That is, if \log_a b = c

    then a^c = b

    LAW 2: \log_a \left(x^n \right) = n \log_a x .....that is, we can make the power of something being logged into a multiplyer

    LAW 3: \log_a xy = \log_a x + \log_a y

    LAW 4: \log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y




    Now on to your question.

    We have \log a = x \mbox { and } \log b = y

    \Rightarrow \log a \sqrt {b} = \log a + \log \sqrt {b} ......LAW 3

    = \log a + \log (b)^{ \frac {1}{2}} .......rewrite the second log

    = \log a + \frac {1}{2} \log b .............apply LAW 2 to second log

    = x + \frac {1}{2}y ..............inserted the values given

    this is choice 4
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    ooh i got stuck with the square root i didnt know you had to make it into an exponent
    for the same regents i also have questions with 16, 17, and 18 if you dont mind helping.
    i did the first part of the regents and these are pretty much the questions i didnt understand. my teacher doesnt really help out too much
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    16. If \left ( a^x \right ) ^{2/3} = \frac{1}{a^2} what is the value of x?
    Two things you need to know about exponents:
    (a^m)^n= a^{nm}
    and
    \frac{1}{a^n} = a^{-n}

    So what this question is saying is:
    a^{2x/3} = a^{-2}

    I can go over the proof if you need me to, but it should be pretty obvious that
    \frac{2x}{3} = -2

    2x = -6

    x = -3

    -Dan
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    What is the third term in the expansion of (cos(x) + 3)^5?
    The binomial theorem says:
    (a + b) = a^n + [ _nC_1 ]a^{n - 1}b^1 + [ _nC_2 ]a^{n-2}b^2 + ... + [ _nC_{n-1} ]a^1b^{n-1} + b^n
    Where  _nC_r = \frac{n!}{r!(n - r)!}
    (Where n is a positive integer.)

    Here a = cos(x), b = 3, and n = 5 so the third term is:
     _5C_2 a^3b^2 = \frac{5!}{2!3!} \cdot cos^3(x) \cdot 3^2 = 90cos^3(x)

    -Dan
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    Quote Originally Posted by topsquark View Post
    Two things you need to know about exponents:
    (a^m)^n= a^{nm}
    and
    \frac{1}{a^n} = a^{-n}

    So what this question is saying is:
    a^{2x/3} = a^{-2}

    I can go over the proof if you need me to, but it should be pretty obvious that
    \frac{2x}{3} = -2

    2x = -6

    x = -3

    -Dan
    so you do the 2x/3=/-2 since they have the same bases of a?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shanegoeswapow View Post
    ooh i got stuck with the square root i didnt know you had to make it into an exponent
    for the same regents i also have questions with 16, 17, and 18 if you dont mind helping.
    i did the first part of the regents and these are pretty much the questions i didnt understand. my teacher doesnt really help out too much
    (16)

    recall that when we raise a number with a power to another power, we multiply the powers to get the new power of the base.

    also recall that 1 over something to a power, means the something to the negative of that power. negative powers means inverse. you should look up all the log and power rules. earboth posted them in a PDF here

    \left( a^x \right)^{ \frac {2}{3}} = \frac {1}{a^2}

    \Rightarrow a^{ \frac {2x}{3}} = a^{-2}

    Now equate the powers since the bases are the same

    \Rightarrow \frac {2x}{3} = -2

    \Rightarrow x = -3


    (17)
    By the Binomial theorem, or Pascal's triangle (you should look these up, either here or at wikipedia)

    The third term of (\cos x + 3 )^5 is 10 \cos^3 x \cdot 3^2 = 90 \cos^3 x


    (18)
    By inspection we see that choice 3 is what we should choose. why?

    x ( x + 6) = -10

    \Rightarrow x^2 + 6x = -10

    \Rightarrow x^2 + 6x + 10 = 0

    \Rightarrow (x + 3)^2 + 1 = 0

    This number is never zero, so the roots are imaginary.

    Why did i decide to choose this choice? if you see something squared equals a negative number, chances are that the guy you want.
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    thanks for the answers, they are very helpful
    i have a quick question for number 7, when you put the values on the calculator then you press stat, which button should you press to get the mean?
    on number 27 i have a faint idea of what to do but im not too sure. would you make the exponents equal by getting the same base?
    number 24: i dont knkow how i should add the fractions to simplest form
    so far thats as far as i got if you can help with those thanks alot
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shanegoeswapow View Post
    thanks for the answers, they are very helpful
    i have a quick question for number 7, when you put the values on the calculator then you press stat, which button should you press to get the mean?
    for number 7, i'm not familiar with that feature on the calculator. here' how to do the problem other wise.

    the sum of all the frequencies is the total amount of numbers used. now, for each x_i, the corersponding f_i tells you how many times it shows up, since it gives the frequency. so for x_1 = 25 we have f_i = 3, that means 25 shows up 3 times.

    to find the mean, it is the sum of all the numbers divided by the amount of numbers you have. remember when summing the numbers, you need to take their frequency into account. we do this by multiplying each number by its corresponding frequency.

    So, mean = \frac {25(3) + 20(2) + 11(5) + 10(4)}{3 + 2 + 5 + 4} = 15
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    Quote Originally Posted by shanegoeswapow View Post
    so you do the 2x/3=/-2 since they have the same bases of a?
    Yup!

    -Dan
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    Quote Originally Posted by shanegoeswapow View Post
    number 24: i dont knkow how i should add the fractions to simplest form
    \frac {1}{x}  + \frac {1}{x + 3} ......the LCM is the product of the denominators

    = \frac {(x + 3) + x}{x(x + 3)}

    = \frac {2x + 3}{x(x + 3)}

    or \frac {2x + 3}{x^2 + 3x} if you expand the denominator
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    18. Which of the following has imaginary roots?

    I'm going to post this one anyway, even though Jhevon has given you a good answer.

    The roots of a quadratic equation have an imaginary component if the discriminant D = b^2 - 4ac is negative.

    So let's put all of these into standard form and see which ones have negative discriminants:
    x(5 + x) = 8 ==> x^2 + 5x - 8 = 0 ==> D = 5^2 - 4 \cdot 1 \cdot (-8) = 57

    x(5 - x) = -3 ==> x^2 - 5x - 3 = 0 ==> D = (-5)^2 - 4 \cdot 1 \cdot (-3) = 37

    x(x + 6) = -10 ==> x^2 + 6x + 10 = 0[/tex] ==> D = 6^2 - 4 \cdot 1 \cdot 10 = -4

    (2x + 1)(x - 3) = 7 ==> 2x^2 - 5x - 7 = 0[/tex] ==> D = (-5)^2 - 4 \cdot 2 \cdot (-7) = 81

    So we see that #3 has a negative discriminant and thus has an imaginary component. (As Jhevon already told you.)

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by shanegoeswapow View Post
    on number 27 i have a faint idea of what to do but im not too sure. would you make the exponents equal by getting the same base?
    what are you talking about? 27 has nothing to do with exponents

    see the modified diagram below. if you have any questions, say so

    By the Law of Sines, we have:

    \frac {110}{\sin 27} = \frac {AF}{\sin 105}

    i trust you can take it from here
    Attached Thumbnails Attached Thumbnails Math B Regents Help?-bearing.gif  
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