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Mathematics 9 Online
OpenStudy (imterribleatmath):

@Will.H

OpenStudy (will.h):

1st we need to find k 81 million = 76million*e^k4 Solve for k e^4k = 1.065789474 K4= in (1.065789474) K = 0.01592895367 Now substitute that in the exponential form And solve when t= 22 P = 76,000,000e^0.01592895367*22 P=107896272 So I would say if we had to approximate to nearest million then 108 million (108,000,000)

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