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briana.img

  • one year ago

Can someone walk me through this problem? I always have a hard time with this.

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  1. briana.img
    • one year ago
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  2. briana.img
    • one year ago
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    @EmilyF03 excuse me?

  3. briana.img
    • one year ago
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    I tried this but it didn't seem to work for me \[ 8^2=4^2+7^2-2(4)(7)\cos\]

  4. ganeshie8
    • one year ago
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    \[ 8^2=4^2+7^2-2(4)(7)\cos(\theta)\]

  5. ganeshie8
    • one year ago
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    isolate \(\cos(\theta)\) first

  6. briana.img
    • one year ago
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    i ended up with -57=cosT but i don't think that's right

  7. ganeshie8
    • one year ago
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    you should end up with -1 = -56cos(T) right?

  8. ganeshie8
    • one year ago
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    dividing -56 both sides gives you 1/56 = cos(T)

  9. briana.img
    • one year ago
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    don't i need to put it in cos^-1(-56) ?

  10. jim_thompson5910
    • one year ago
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    cos(T) ? shouldn't it be cos(R) since you want to find m < R ?

  11. briana.img
    • one year ago
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    @jim_thompson5910 ooh oops yepp i was just going off an example in my notes

  12. jim_thompson5910
    • one year ago
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    ah, gotcha

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spraguer (Moderator)
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