Answer:
% change in stopping distance = 7.34 %
Step-by-step explanation:
The stooping distance is given by

We will approximate this distance using the relation

dx = 26 - 25 = 1
T' = 2.5 + x
Therefore

This is the stopping distance at x = 25
Put x = 25 in above equation
2.5 × (25) + 0.5×
+ 2.5 + 25 = 402.5 ft
Stopping distance at x = 25
T(25) = 2.5 × (25) + 0.5 × 
T(25) = 375 ft
Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft
% change in stopping distance =
× 100
% change in stopping distance = 7.34 %
Let the side length of the square be x, then A = x^2
but diagonal (z) = sqrt(2x^2)
z^2 = 2x^2
x^2 = 1/2 z^2
Thus, A = 1/2 z^2
dA/dz = 1/2 (2z) = z
The rate of change is z.
When z = 4, the rate is 4.
Answer:
y=54/7
Step-by-step explanation:
Answer:
<h2>
y = 6x - 1</h2>
Step-by-step explanation:

(0, -1) ⇒ x₁ = 0, y₁ = -1
(1, 5) ⇒ x₂ = 1, y₂ = 5
So the slope:

The slope-intercept form of the equation of line is y = mx + b, where m is the slope and b is the y-intercept of the line.
(0, -1) ⇒ x₀ = 0, y₀ = -1 ⇒ b = -1
Therefore:
y = 6x - 1 ← the slope-intercept form of the equation