Tutor profile: Krishna K.
Subject: Linear Algebra
Solve for X 2(X - 1) + X = 5(2X + 3) - 2(X + 3)
Solution: 1) Expand brackets 2X - 2 + X = 10X + 15 - 2X - 6 1) Isolate Variable X 2X + X - 10X + 2X = 2 + 15 - 6 3) Simplify -5X = 11 4) Divide by -5 X = -11/5
A vehicle travels at a distance of d(t) = (50 x t^2) meters. What is the acceleration of the vehicle?
Acceleration is the 2nd derivative of distance. d(t) = (50 x t^2) meters d'(t) = v(t) = (100 x t) meters/second d''(t) = a(t) = v'(t) = (100) meters/second^2 The acceleration of the vehicle is 100 m/s^2
Subject: Basic Math
Sam and Jerry are planning to meet up to chat for 10 mins during lunch. They both work 5 km apart from each other and can only take a 1-hour lunch break, or else they will be late. Sam can travel using his bicycle going 10km/hour. Jerry can travel by foot going 3km/hour. If both Sam and Jerry leave at 12:00PM, What time will Jerry return back to the office? Will he be late?
Information Extracted: - 10 min chat once they meet - 5 km distance - 1 hour max time - Sam: 10km/h - Jerry: 3km/h Variables Used: (T) = Time of when Sam and Jerry meet (hours) (S) = Sam's Distance to Jerry (J) = Jerry's Distance to Sam Create these 2 equation using what we know already. 1. (S) = (T) x (10km/hour) 2. (J) = (T) x (3km//hour) 3. (S + J) = 5km Develop new equations using a common variable (T). 4. (T) = (S) / (10km/hour) 5. (T) = (J) / (3km/hour) 6. (S) = (T) x (10km/hour) 7. (J) = (T) x (3km/hour) Add equation 6&7 7. (T) x (10km/hour + 3km/hour) = (S + J) 8. (T) x (13km/hour) = 5km We can solve for T. T = (5km) / (13km/hour) = 0.38 hours Jerry will return back to the office after he meets with Jerry (0.38 hours), chat with him (10 mins) and walk back to the office (0.38 hours). Total Trip Time = 0.38 hours + 0.38 hours + 0.166 hours = 0.92 hours Time Arrival back to office = (0.92 hours * 60 mins) = 55.6 mins Therefore Jerry will return back to the office at 12:56 PM and he will not be late.
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