Solve 2 2 2 + + (22 − 2) = 0, ≠ 0, ≥ 0
1 answer:
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Answer:
Step-by-step explanation:
Let v be the speed after the collision of both the cars.
Now, momentum is given by: mass × velocity, then by equating the total momentum before and after the collision of the two cars, we get
⇒
⇒
⇒
Thus, they are moving at the velocity of
The vertex (minimum) of the quadratic ax² +bx +c is located at x=-b/(2a). This means the minimum value of f(x) will be found at x = -3/(2*1) = -1.5.
Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.
On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057
Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.
On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057
Answer:
y = x + 1
Step-by-step explanation:
the equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
calculate m using the slope formula
m =
with (x₁, y₁ ) = (- 2, 0) and (x₂, y₂ ) = (0, 1) ← 2 points on the line
m = =
=
the line crosses the y- axis at (0, 1 ) ⇒ c = 1
y = x + 1 ← equation of line