Use the rules of logarithms and the rules of exponents.
... ln(ab) = ln(a) + ln(b)
... e^ln(a) = a
... (a^b)·(a^c) = a^(b+c)
_____
1) Use the second rule and take the antilog.
... e^ln(x) = x = e^(5.6 + ln(7.5))
... x = (e^5.6)·(e^ln(7.5)) . . . . . . use the rule of exponents
... x = 7.5·e^5.6 . . . . . . . . . . . . use the second rule of logarithms
... x ≈ 2028.2 . . . . . . . . . . . . . use your calculator (could do this after the 1st step)
2) Similar to the previous problem, except base-10 logs are involved.
... x = 10^(5.6 -log(7.5)) . . . . . take the antilog. Could evaluate now.
... = (1/7.5)·10^5.6 . . . . . . . . . . of course, 10^(-log(7.5)) = 7.5^-1 = 1/7.5
... x ≈ 53,080.96
Answer: The approximate absentee rate that day would be 8.09%.
Step-by-step explanation:
Since we have given that
Number of students who were absent = 36
Total number of students = 445
We need to find the approximate absentee rate that day :
Rate of absentee of that day would be
Hence, the approximate absentee rate that day would be 8.09%.
H(x) = 6x
it gives you what x is so plug that in the equation to find it.
h(2/3) = 6(2/3)
h(2/3) = 6 × 2 ÷ 3
h(2/3) = 12 ÷ 3
h(2/3) = 4
so your answer is 4.
hope this helps, God bless!
Answer:
16.5
Step-by-step explanation:
this is very true because it is
You replace the y with f(x)
So it becomes f(x)=2x+1
