Answer:
It is expected that linearization beyond age 20 will be use a function whose slope is monotonously decreasing.
Step-by-step explanation:
The linearization of the data by first order polynomials may be reasonable for the set of values of age between ages from 5 to 15 years, but it is inadequate beyond, since the fourth point, located at 
, in growing at a lower slope. It is expected that function will be monotonously decreasing and we need to use models alternative to first order polynomials as either second order polynomic models or exponential models.
Remember:
a^m * a^n=a^(m+n)
a^-m=1/a^m
8³ x 8⁻⁵=8³⁻⁵=8⁻²=1/8²=1/64<1
Answer: the student is not correct, because 1/64<1, and 8³ x 8⁻⁵=1/64.
:) Brainliest pls?
Answer:
The zeros are {2, -3, -4} which need to be plotted on the x-axis.
Step-by-step explanation:
I'll find the zeros, aka x-intercepts, and you could probably graph them.
To find the zeros, let's factor this polynomial:
r(x) = (x - 2)(x^2+7x+12)
r(x) = (x - 2)(x + 3)(x + 4)
The zeros are {2, -3, -4} which need to be plotted on the x-axis.
3 ln 3 + 4 ln b => ln 3^3 + ln b^4 => ln (3^3)*(b^4)
OH I KNOW I KNOW TEACHER. its its wait im in middle school what the heck am i doing in here.
