Based on the conditions given above, the number of bacteria at any time t (in hours) is calculated by the equation,
at = (a1)(2^t/2)
where a1 is the initial number of bacteria and at is the number at any time t. Substituting the givens,
a6 = (103)(2^6/2) = 824
Thus, there are 824 bacteria after 6 hours.
Answer:
no he will make it home at 18:15
Step-by-step explanation:
Hey there!
-1.3n = 7.8
First, you have to DIVIDE both sides by -1.3
-1.3n / -1.3 = 7.8 / -1.3
Cancel out: -1.3 / -1.3 because it gives you the value of 1
Keep: 7.8 / -1.3 because it gives us the value of n
7.8 ÷ -1.3 = -6
Thus, n = -6 ✅
Good luck on your assignment and enjoy your day!
~LoveYourselfFirst:)
This can be solve by using the formula
D = P( 1 – i)^n
Where d is the depreciation value after n years
P is the initial value
i is the depreciation rate
n is the years
D = 1/3 ( 1800)
D = 600
So
600 = 1800 ( 1- 0.45)^n
Solve for n
<span>N = 1.83 years</span>
Answer:
All real number
Step-by-step explanation:
The graph of y = x is shown to the right. We can see that it is defined for all x-values because the line will continue to infinity. Thus, the domain of y = x is all real numbers.
Hope this help <3
