I believe that the answer is C
Let y = 12 e^2x
e^2x = y/12
Taking
[email protected]ln e^2x = ln (y/12)
2x = ln (y/12)
x = (1/2) ln (y / 12)
so the inverse h-1(x) = (1/2) ln ( x / 12)
The correct numbers to use in solving problems about
spans of time like B.C. and A.D. should be “integers”.
Integers are whole numbers (not a fractional number or not a decimal
number) which can take a value of negative, zero, or positive number. Example
of integers would be -1, 0 and 1.
<span>In calculations, the time period would be on the x-axis. Since
B.C. and A.D. are two different spans of time, therefore in the calculations,
the date of BC should be negative (negative x-axis) while the date of AD should
be positive (positive x-axis). This would place the origin as the common
reference.</span>
Any value of y, except when y = -4.9, will make the expression equal to 2.45.
The function is computed below based on the information. The answers will be:
1. -11
2. -1
<h3>How to compute the function?</h3>
f(x) = x²-11x+28
g(x) = x-4
(f+g)(-5) = (-5² - 11 × 5 + 28) + (-5 - 4)
= 25 - 55 + 28 - 9
= -11
(f-g)(5)
= = (5² - 11 × 5 + 28) + (5 - 4)
= 25 - 55 + 28 + 1
= -1
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