Let's define variables:
s = original speed
s + 12 = faster speed
The time for the half of the route is:
60 / s
The time for the second half of the route is:
60 / (s + 12)
The equation for the time of the trip is:
60 / s + 60 / (s + 12) + 1/6 = 120 / s
Where,
1/6: held up for 10 minutes (in hours).
Rewriting the equation we have:
6s (60) + s (s + 12) = 60 * 6 (s + 12)
360s + s ^ 2 + 12s = 360s + 4320
s ^ 2 + 12s = 4320
s ^ 2 + 12s - 4320 = 0
We factor the equation:
(s + 72) (s-60) = 0
We take the positive root so that the problem makes physical sense.
s = 60 Km / h
Answer:
The original speed of the train before it was held up is:
s = 60 Km / h
Solve out
x^2 + 64 = 0
x^2 = -64 ( we are going to use imaginary numbers to solve)
x =
x = 8i, -8i final answers
Answer:
Make sure the mode is on degree and not radian! Common mistake
<span>In order to determine the value of the numbers, we can set up algebraic equations to solve. For this case, we need to set up two equations since we have two unknown numbers. We do as follows:
let x = first number and y = second number
From the problem statement, we set up equations.
Equation 1 - the numbers have a difference of 0.7
x - y = 0.7
Equation 2 - the numbers have a sum of 1
x + y = 1
Solving for x and y via substitution method,
x - y = 0.7
(1-y) - y = 0.7
1 - 2y = 0.7
-2y = 0.7 - 1 = -0.3
y = 0.15 or 3/20
x - 0.15 = 0.7
x = 0.85 or 17/20
Thus, the two numbers are 0.15 and 0.85.</span>
Answer:
B
Step-by-step explanation:
(a+b)^2
using Pascal triangle
1 1
1 2 1
1 3 3 1
so the third term in the expansion of (a+b)^2
we use
1 2 1
a^2 + 2ab+b
hence the coefficient of third term is 1
option B is collect
