Answer:
C
Step-by-step explanation:
Notice how the angles look the same, but if we look at the base of our second house, it's value is triple that of the first house. Therefore, we can infer that the other values must be triple the size, as well. Thus, multiplying our values of the first house, we get answer C!
Hope this helped!
Here's the general formula for bacteria growth/decay problems
Af = Ai (e^kt)
where:
Af = Final amount
Ai = Initial amount
k = growth rate constant
<span>t = time
But there's another formula for a doubling problem.
</span>kt = ln(2)
So, Colby (1)
k1A = ln(2) / t
k1A = ln(2) / 2 = 0.34657 per hour.
So, Jaquan (2)
k2A = ln(2) / t
<span>k2A = ln(2) /3 = 0.23105 per hour.
</span>
We need to use the rate of Colby and Jaquan in order to get the final amount in 1 day or 24 hours.
Af1 = 50(e^0.34657(24))
Af1 = 204,800
Af2 = 204,800 = Ai2(e^0.23105(24))
<span>Af2 = 800</span>
Answer:
The 1st problem is C. 212.79 The 2nd problem is 1 3/5 The 3rd problem is 3.68
Step-by-step explanation:
Explanation for problem 1 l times w = area
Explanation for problem 2 change to decimals multiply w times l change back to fraction
Explanation for problem 3 multiply l times w
Answer:
what grade is this?
Step-by-step explanation:
I believe the answer is 40%
Not entirely sure, hope I helped though!
