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2 years ago

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A jogger started a course and jogged at an average speed of 4 mph. One hour later, a cyclist started the same course and cycled

at an average speed of 11 mph. How long after the jogger started did the cyclists overtake the jogger.

1 answer:

Anna007 [38]2 years ago

6 0

4/(11-4)=4/7hours :D

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$\bf \cfrac{\stackrel{\textit{perfect square trinomial}}{cos^2(\theta )+2cos(\theta )+1}}{sin^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the right-hand-side}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}}\implies \stackrel{\textit{multiplying by the denominator's conjugate}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}\cdot \cfrac{1+cos(\theta )}{1+cos(\theta )}}$

$\bf \cfrac{[1+cos(\theta )]^2}{\underset{\textit{difference of squares}}{[1-cos(\theta )][1+cos(\theta )]}}\implies \cfrac{[cos(\theta )+1]^2}{1^2-cos^2(\theta )} \\\\\\ \cfrac{[cos(\theta )+1]^2}{1-cos^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}}$

recall that sin²(θ) + cos²(θ) = 1, thus sin²(θ) = 1 - cos²(θ).

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