Answer:
Input Data :
Point 1(xA,yA)(xA,yA) = (4, 3)
Point 2(xB,yB)(xB,yB) = (3, -2)
Objective :
Find the distance between two given points on a line?
Formula :
Distance between two points = √(xB−xA)2+(yB−yA)2(xB-xA)2+(yB-yA)2
Solution :
Distance between two points = √(3−4)2+(−2−3)2(3-4)2+(-2-3)2
= √(−1)2+(−5)2(-1)2+(-5)2
= √1+251+25
= √2626 = 5.099
Distance between points (4, 3) and (3, -2) is 5.099
Answer:
General Formulas and Concepts:
<u>Calculus</u>
Limits
Limit Rule [Variable Direct Substitution]:
L'Hopital's Rule
Differentiation
Basic Power Rule:
Derivative Rule [Chain Rule]:
Step-by-step explanation:
We are given the limit:
When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:
This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:
Plugging in <em>x</em> = 0 again, we would get:
Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:
Substitute in <em>x</em> = 0 once more:
And we have our final answer.
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Limits