anonymous
  • anonymous
NVM
Mathematics
chestercat
  • chestercat
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whpalmer4
  • whpalmer4
For problem s, you can see if you do a little bit of work that the two sides are the same length, and the sum of the squares of their lengths is the square of the length of the hypotenuse. Therefore the answer choice involving the Pythagorean theorem is correct. Here's how I worked it out: The two sides both go 7 in one direction and 1 in the other, so we can find the length by the Pythagorean theorem: \(d = \sqrt{7^2 + 1^2} = \sqrt{50} = 5\sqrt{2}\) The hypotenuse goes 8 units in one direction and 6 in the other, so its length is:\[d=\sqrt{8^2+6^2} = \sqrt{100} = 10\] We can see that these lengths satisfy the Pythagorean theorem: \((5\sqrt{2})^2 + (5\sqrt{2})^2 = 25*2 + 25*2 = 50+50 = 10^2 = 100\) So, this is a right triangle (and an isosceles one, at that).
whpalmer4
  • whpalmer4
In #9 in file h, remember the memory aid: SOH CAH TOA Sin = Opposite over Hypotenuse Cos = Adjacent over Hypotenuse Tan = Opposite over Adjacent
whpalmer4
  • whpalmer4
You'll have to give it your best shot...

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