Greetings from Brasil...
F(V) = - 11V² - 8
F(a + h) - F(a)]/h
let's do it by steps
F(a + h) = - 11 · (a + h)² - 8
F(a + h) = - 11 · (a² + 2ah + h²) - 8
F(a + h) = - 11a² - 22ah - 11h² - 8
<em>and</em>
F(a) = - 11 · (a)² - 8
F(a) = - 11a² - 8
Then
F(a + h) - F(a)]/h
{- 11a² - 22ah - 11h² - 8 - [- 11a² - 8]}/h
[- 11a² - 22ah - 11h² - 8 + 11a² + 8]/h
[- 22ah - 11h²]/h
h · [- 22a - 11h]/h
<h2>[- 22a - 11h]</h2>
32 39 46 53 60 67 74 81 88 95 102 do I have to continue?
<h3>
Answer:</h3>
- P(B|A) = the probability you win the second game given that you win the first game
<h3>
Step-by-step explanation:</h3>
The notation P(B|A) means "the probability of event B given event A". For the two events defined in the problem statement, the expression means ...
... the probability <em>you win the second game</em> given <em>you win the first game</em>.
Answer:
C P(v) = 2(v+7)
Step-by-step explanation:
To find P(v), we need to take the inverse of P^-1 (v)
y = 1/2 v-7
Exchange y and v
v = 1/2 y-7
Solve for y
Add 7 to each side
v+7 = 1/2 y -7+7
v+7 = 1/2y
Multiply each side by 2
2(v+7) = 1/2 y*2
2(v+7) = y
P(v) = 2(v+7)
Answer:
each cd was 1 cent
Step-by-step explanation:
