Solve the equation 6/5y-10=-4 X=
1 answer:
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A)
![\bf 1990-1910=80\leftarrow t \\\\\\ P(t)=8000(2)^{-\frac{t}{29}}\implies P(80)=8000(2)^{-\frac{80}{29}} \\\\\\ P(80)=8000\cdot \cfrac{1}{2^{\frac{80}{29}}}\implies P(80)=\cfrac{8000}{\sqrt[29]{2^{80}}}](https://tex.z-dn.net/?f=%5Cbf%201990-1910%3D80%5Cleftarrow%20t%0A%5C%5C%5C%5C%5C%5C%0AP%28t%29%3D8000%282%29%5E%7B-%5Cfrac%7Bt%7D%7B29%7D%7D%5Cimplies%20P%2880%29%3D8000%282%29%5E%7B-%5Cfrac%7B80%7D%7B29%7D%7D%0A%5C%5C%5C%5C%5C%5C%0AP%2880%29%3D8000%5Ccdot%20%5Ccfrac%7B1%7D%7B2%5E%7B%5Cfrac%7B80%7D%7B29%7D%7D%7D%5Cimplies%20P%2880%29%3D%5Ccfrac%7B8000%7D%7B%5Csqrt%5B29%5D%7B2%5E%7B80%7D%7D%7D)
and surely you know how much that is.
b)
since in 1910 t = 0, 174 years later from 1910, is 2084, so in 2084 they'll be 125 exactly, so the next year, 2085, will then be the first year they'd fall under that.
and surely you know how much that is.
b)
since in 1910 t = 0, 174 years later from 1910, is 2084, so in 2084 they'll be 125 exactly, so the next year, 2085, will then be the first year they'd fall under that.