Answer:
each classmate gets half of a pizza
Answers is 1000 in. Hope this helped
B - 4.2 < -7.5
subtract b and 4.2
less then sign is <
then <-7.5
The fare of $(20 - 2.5) = $17.5 will maximize the total fare.
<h3>What is Differentiation?</h3>
Differentiation means the rate of change of one quantity with respect to another. The speed is calculated as the rate of change of distance with respect to time.
Here, The operator for a round-trip fare of $20, carries an average of 500 people per day.
It is estimated that 20 fewer people will take the trip, for each $1 increase in fare.
for $x increase in fare, 20x less people will take the trip and at that time the total fare F is given by
f(x) =(20 + x)(500 - 20x)
f (x) = 10000 + 100x - 20x²
For f(x) to be maximum, the condition is dy/dx = 0
100 - 40x = 0
⇒ x = 2.5
Thus, the fare of $(20 - 2.5) = $17.5 will maximize the total fare.
Learn more about Differentiation from:
brainly.com/question/24062595
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<h3>
Answer: Choice A</h3>
Domain = (a,b]
Range = [mc + n,md + n)
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Explanation:
The domain stays the same because we still have to go through f(x) as our first hurdle in order to get g(x).
Think of it like having 2 doors. The first door is f(x) and the second is g(x). The fact g(x) is dependent on f(x) means that whatever input restrictions are on f, also apply on g as well. So going back to the "2 doors" example, we could have a problem like trying to move a piece of furniture through them and we'd have to be concerned about the f(x) door.
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The range will be different however. The smallest value in the range of f(x) is y = c as it is the left endpoint. So the smallest f(x) can be is c. This means the smallest g(x) can be is...
g(x) = m*f(x) + n
g(x) = m*c + n
All we're doing is replacing f with c.
So that means mc+n is the starting point of the range for g(x).
The ending point of the range is md+n for similar reasons. Instead of 'c', we're dealing with 'd' this time. The curved parenthesis says we don't actually include this value in the range. A square bracket means include that value.