![\bf \lim\limits_{x\to \infty}~\left( \cfrac{1}{8} \right)^x\implies \lim\limits_{x\to \infty}~\cfrac{1^x}{8^x}\\\\[-0.35em] ~\dotfill\\\\ \stackrel{x = 10}{\cfrac{1^{10}}{8^{10}}}\implies \cfrac{1}{8^{10}}~~,~~ \stackrel{x = 1000}{\cfrac{1^{1000}}{8^{1000}}}\implies \cfrac{1}{8^{1000}}~~,~~ \stackrel{x = 100000000}{\cfrac{1^{100000000}}{8^{100000000}}}\implies \cfrac{1}{8^{100000000}}~~,~~ ...](https://tex.z-dn.net/?f=%5Cbf%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D~%5Cleft%28%20%5Ccfrac%7B1%7D%7B8%7D%20%5Cright%29%5Ex%5Cimplies%20%5Clim%5Climits_%7Bx%5Cto%20%5Cinfty%7D~%5Ccfrac%7B1%5Ex%7D%7B8%5Ex%7D%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7Bx%20%3D%2010%7D%7B%5Ccfrac%7B1%5E%7B10%7D%7D%7B8%5E%7B10%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%5E%7B10%7D%7D~~%2C~~%20%5Cstackrel%7Bx%20%3D%201000%7D%7B%5Ccfrac%7B1%5E%7B1000%7D%7D%7B8%5E%7B1000%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%5E%7B1000%7D%7D~~%2C~~%20%5Cstackrel%7Bx%20%3D%20100000000%7D%7B%5Ccfrac%7B1%5E%7B100000000%7D%7D%7B8%5E%7B100000000%7D%7D%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%5E%7B100000000%7D%7D~~%2C~~%20...)
now, if we look at the values as "x" races fast towards ∞, we can as you see above, use the values of 10, 1000, 100000000 and so on, as the value above oddly enough remains at 1, it could have been smaller but it's constantly 1 in this case, the value at the bottom is ever becoming a larger and larger denominator.
let's recall that the larger the denominator, the smaller the fraction, so the expression is ever going towards a tiny and tinier and really tinier fraction, a fraction that is ever approaching 0.
Answer:
68°F is 20°C
Step-by-step explanation:
C = (5/9)(F - 32)
C = (5/9)(68 - 32)
C = (5/9)(36)
C = 5 * 4
C = 20
Answer: 68°F is 20°C
Assuming you're working in base 10, then the antilog of 3.493 is equal to 10^3.493 = 3,111.71633710601
This rounds to 3111.7163 so the answer is choice D
Answer:
the father ends up paying 18 dollars after everything