Answer:
Extraneous Solutions An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. Example 1: Solve for x, 1 x − 2 + 1 x + 2 = 4 (x − 2) (x + 2).
Log(2)/log(1.064) ≈ 11.17 . . . . hours
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The population can be given by
p(n) = p₀×1.064ⁿ . . . . where n is the number of hours
You want to find n whe p(n) = 2*p₀.
2p₀ = p₀×1.064ⁿ . . . . . . . . . . . . substitute the given information
2 = 1.064ⁿ . . . . . . . . . . . . . . . . . divide by p₀
log(2) = n×log(1.064) . . . . . . . . take logs to make it a linear equation
log(2)/log(1.064) = n . . . . . . . . divide by the coefficient of n
Answer:
157.1 m
Step-by-step explanation:
Here, we want to calculate the length of the semi-circle
This simply refers to the circumference of the semi-circle
Mathematically, the circumference is half that of a full circle
The circumference of a full circle is;
C = pi * d
For half circle or semi-circle;
It is ;
C = (pi * d)/2
D = 50 m
The circumference is thus;
C = 50 * 3.142
C = 157.1 m
Answer:
s=6
Step-by-step explanation:
72/12=6 simplify both sides
