~ Simplifying
-2(7x + -5) = -14x + 10
~ Reorder the terms:
-2(-5 + 7x) = -14x + 10
(-5 * -2 + 7x * -2) = -14x + 10
(10 + -14x) = -14x + 10
~ Reorder the terms:
10 + -14x = 10 + -14x
~ Add '-10' to each side of the equation.
10 + -10 + -14x = 10 + -10 + -14x
~ Combine like terms: 10 + -10 = 0
0 + -14x = 10 + -10 + -14x
-14x = 10 + -10 + -14x
~ Combine like terms: 10 + -10 = 0
-14x = 0 + -14x
-14x = -14x
~ Add '14x' to each side of the equation.
-14x + 14x = -14x + 14x
~ Combine like terms: -14x + 14x = 0
0 = -14x + 14x
~ Combine like terms: -14x + 14x = 0
0 = 0
~ Solving
0 = 0
~ This equation is an identity, all real numbers are solutions.
Answer:
The data we have is:
The acceleration is 3.2 m/s^2 for 14 seconds
Initial velocity = 5.1 m/s
initial position = 0m
Then:
A(t) = 3.2m/s^2
To have the velocity, we integrate over time, and the constant of integration will be equal to the initial velocity.
V(t) = (3.2m/s^2)*t + 5.1 m/s
To have the position equation, we integrate again over time, and now the constant of integration will be the initial position (that is zero)
P(t) = (1/2)*(3.2 m/s^2)*t^2 + 5.1m/s*t
Now, the final position refers to the position when the car stops accelerating, this is at t = 14s.
P(14s) = (1/2)*(3.2 m/s^2)*(14s)^2 + 5.1m/s*14s = 385m
So the final position is 385 meters ahead the initial position.
Answer:
13.5%
Step-by-step explanation:
70 = 7 x 2 x 5
60 = 2² x 3 x 5
50 = 2 x 5²
LCM = 2 x 3 x 5 x 7 = 210
Answer: LCM = 210
