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Math Help - some more problems!!

  1. #1
    Junior Member
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    Red face some more problems!!

    i have not able to answer any of that at all , i feel that i will never pass the final year exam with all that stress on my head ...need please ,but only if u guys know any them .
    Attached Thumbnails Attached Thumbnails some more problems!!-untitled-1.jpg   some more problems!!-untitled-2.jpg   some more problems!!-untitled-3.jpg  
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  2. #2
    Senior Member ecMathGeek's Avatar
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    1. Find  \frac {dy}{dx} at x=2 if y=2e^x+x^3

    y^\prime(x)=2e^x + 3x^2

    Plugging in x=2 we get:

    y^\prime(2)=2e^{(2)}+3(2)^2=2e^2+12

    Do you need this explained more?

    I'm not sure what method you need to use to solve number 2 of the first page. Are you expected to use Newton's method or something else?
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    1. Find  \frac {dy}{dx} at x=2 if y=2e^x+x^3

    y^\prime(x)=2e^x + 3x^2

    Plugging in x=2 we get:

    y^\prime(2)=2e^{(2)}+3(2)^2=2e^2+12

    Do you need this explained more?
    i see you're getting the hang of LaTex already. I still use ' instead of typing ^\prime though, it looks the same
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  4. #4
    Senior Member ecMathGeek's Avatar
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    3. a) Prove that 9cos(x)+6sin(x)=10

    Notice that  \angle QBA, \angle ABC and \angle CBR are suplementary (they sum up to equal 180^\circ), and that \angle ABC=90^\circ. Therefore, \angle QBA and \angle CBR are complementary. This means that \angle CBR \cong \angle QAB and thus that \triangle AQB \cong \triangle BRC.

    With all of that said. Here's where I can start the proof:

    Notice that  x = \angle CBR = \angle QAB, Therefore: sin(x) = \frac {QB}{AB} and  cos(x)=\frac {BR}{BC}. However,  QB = 30 - BR. Therefore:

     sin(x) = \frac {30-BR}{18} \rightarrow 18sin(x) = 30-BR

    cos(x) = \frac {BR}{27} \rightarrow 27cos(x) = BR

    If we add these two equations, we get:

     18sin(x) + 27cos(x) = 30 -BR+BR=30

    Now divide everything by 3 to get:

     6sin(x) + 9cos(x) = 10
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Jhevon View Post
    i see you're getting the hang of LaTex already. I still use ' instead of typing ^\prime though, it looks the same
    I'm learning it... and still making plenty of mistakes. But it's a nice program.
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  6. #6
    Senior Member ecMathGeek's Avatar
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    3. b) Express 9cos(x)+6sin(x) in the form Rcos(x-a) and find the values of R and a to 2 decimal places.

    Notice that Rcos(x-a)=Rcos(x)cos(a)+Rsin(x)sin(a) but we want 9cos(x)+6sin(x)=Rcos(x-a). So we want 9cos(x)+6sin(x)=Rcos(x)cos(a)+Rsin(x)sin(a). Therefore, we need 9=Rcos(a) and 6=Rsin(a).

    To solve this will take some creativity.

    We have:
    9=Rcos(a) \rightarrow cos(a) = \frac {9}{R}
    6=Rsin(a) \rightarrow sin(a) = \frac {6}{R}

    Let's square both sides of each equation:
    cos^2(a)= \frac {81}{R^2}
    sin^2(a)= \frac {36}{R^2}

    Now, we'll use the identity cos^2(a) = 1-sin^2(a) to get:
     1-sin^2(a) = \frac {81}{R^2} \rightarrow sin^2(a) = 1-\frac{81}{R^2}

    Setting the two sin^2(a) equal, we get:

     \frac {36}{R^2} = 1-\frac{81}{R^2}

     36 = R^2-81 \rightarrow R^2 = 117 \rightarrow R \approx 10.82

    Using this, we can find a:

    cos(a) = \frac {9}{10.817}

    a = cos^{-1}(\frac{9}{10.817}) \rightarrow a \approx 33.69
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  7. #7
    Senior Member ecMathGeek's Avatar
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    3. c) Find the value of x.

    10.82cos(x-33.69)=10
    cos(x-33.69)=\frac {10}{10.82}
    x-33.69= cos^{-1}(\frac {10}{10.82})
    x = 33.69+cos^{-1}(\frac {10}{10.82})
    x= 56.1
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