The average rate of change of a function, g(x), over an interval [a, b] is given by
Thus, the average rate of change of
over the interval
is given by:
![Average\ rate\ of\ change= \frac{[-(8)^2-4(8)]-[-(6)^2-4(6)]}{8-6} \\ \\ = \frac{(-64-32)-(-36-24)}{2} = \frac{-96-(-60)}{2} = \frac{-96+60}{2} = \frac{-36}{2} \\ \\ -18](https://tex.z-dn.net/?f=Average%5C%20rate%5C%20of%5C%20change%3D%20%5Cfrac%7B%5B-%288%29%5E2-4%288%29%5D-%5B-%286%29%5E2-4%286%29%5D%7D%7B8-6%7D%20%5C%5C%20%20%5C%5C%20%3D%20%5Cfrac%7B%28-64-32%29-%28-36-24%29%7D%7B2%7D%20%3D%20%5Cfrac%7B-96-%28-60%29%7D%7B2%7D%20%3D%20%5Cfrac%7B-96%2B60%7D%7B2%7D%20%3D%20%5Cfrac%7B-36%7D%7B2%7D%20%20%5C%5C%20%20%5C%5C%20-18)
Answer:
Okay I'm not thoroughly understanding your question. But here is my answer for BOTH ways.
Step-by-step explanation:
If it's 5 TIMES 3x/8 +2=7
x= 8/3 or x=2.6
If it's 5(3x) OVER 8 in fraction form +2=7
x= 11
Hope this answers your question.
After solving the provided question, Aisha spent 7.9 hours and 9.9 hours, respectively, working last week as a babysitter and a landscaper.
<h3>Which is an equation?</h3>
A mathematical statement that has at least two terms, both of which contain variables or equal integers, is defined as an equation.
Aisha is reportedly working two summer jobs, earning $7 an hour for babysitting and $8 an hour for gardening.
Let's say you were to watch the children for x hours.
And y would be the amount of hours spent on landscaping.
Last week, Aisha spent two more hours babysitting than a landscaper, earning a total of $134.
To more about the equation here:
brainly.com/question/649785
#SPJ1
Step-by-step explanation:
An exponentially decaying signal is of the form x(t) = Ce^(-αt) in terms of an initial value C and a decay rate α > 0. The signal equals a fraction 1/e of its initial value after the characteristic time scale t = 1/α.
Given
x(t) = 2e^(-t/3) + e^(-t) + 3e^(-t/2)
The decay rates are: 1/3, 1, and 1/2.
The slowest decay rate α is the minimum of {1/3, 1, 1/2} = 1/3.
The corresponding time scale
is only 2/3 times larger than the next faster decay rate 1/2, so the decay rates are NOT well separated. The slowest component is larger than 0.5 as long as t < (ln3)/0.5 ≈ 2.2.
The other two components add
up to more than the value of the slowest component. We conclude that the component with slowest decay rate dominates measurements on any interval even in the presence of noise.
Measure = 1/2 (120-30)
1/2(90)
45
Answer = 45°
