Given: ∠AOB is a central angle and ∠ACB is a circumscribed angle. Prove: △ACO ≅ △BCO Circle O is shown. Line segments A O and B
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When it comes to laplace equations, there are transformation equations to follow. Generally, when you want to transform a laplace equation, you change the equation from f(t) to F(s). If you do the reverse, it is called the reverse laplace equation.
Based on the given, the useful transformation equation is shown in the attached picture.
When the term is s^2, that must mean that the equation is 1!/s^(1+1) to yield 1/s^2. This means that n=1. Taking the reciprocal s^2 must be equal to 1/t. Thus, for the first term, -11s^2 is equal to -11/t. For the second term, n must be equal to 6 so that 6!/s^(6+1) would yield 720/s^7. Thus, 720s^7 is equal to 1/t^6.
Hence, the transformed equation is -11/t - 1/t^6
Based on the given, the useful transformation equation is shown in the attached picture.
When the term is s^2, that must mean that the equation is 1!/s^(1+1) to yield 1/s^2. This means that n=1. Taking the reciprocal s^2 must be equal to 1/t. Thus, for the first term, -11s^2 is equal to -11/t. For the second term, n must be equal to 6 so that 6!/s^(6+1) would yield 720/s^7. Thus, 720s^7 is equal to 1/t^6.
Hence, the transformed equation is -11/t - 1/t^6