Which 2 storage have similar shapes?

The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.

So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.

Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.

The others must end in 3.

They must be greater than 152 which is 225. So the

First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.

The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.

7 0

3 years ago

Assume you have such a watch. if a minimum of 18.0% of the original tritium is needed to read the dial in dark places, for how m

laila [671]

Given that the half life of 3H is 12.3 years.

The amount of substance left of a radioactive substance with half life of $t_{ \frac{1}{2} }$ after t years is given by

$N(t)=N_0\left(\frac{1}{2} \right)^{ \frac{t}{t_{\frac{1}{2}} }$

Therefore, the number of years it will take for 18% of the original tritinum to remain is given by

$18 \% = 100 \% \left(\frac{1}{2} \right)^{ \frac{t}{12.3}} \\ \\ \Rightarrow\left(\frac{1}{2} \right)^{ \frac{t}{12.3}} =0.18 \\ \\ \Rightarrow\frac{t}{12.3}\ln\left(\frac{1}{2} \right)=\ln0.18 \\ \\ \Rightarrow\frac{t}{12.3}= \frac{\ln0.18}{\ln\left(\frac{1}{2} \right)} = \frac{-1.715}{-0.6931} =2.474\\ \\ \Rightarrow t=2.474(12.3)=30.4$

Therefore, the number of <span>years that the time could be read at night is 30.4 years.</span>

5 0

3 years ago