<em>v₂ = 55mph</em>
- Step-by-step explanation:
<em>Hi there ! </em>
<em>Velocity formula</em>
<em>v = distance/time = d/t => d = vₓt</em>
<em>v₁ = v₂ - 11</em>
<em>t₁ = 10h</em>
<em>t₂ = 8h</em>
<em>d = v₁ₓt₁ = (v₂ - 11)t₁ </em>
<em>d = v₂t₂</em>
<em>(v₂ - 11)t₁ = v₂t₂</em>
<em>replace t₁ ; t₂</em>
<em>(v₂ - 11)10 = v₂8</em>
<em>10v₂ - 11ₓ10 = 8v₂</em>
<em>10v₂ - 8v₂ = 110</em>
<em>2v₂ = 110</em>
<em>v₂ = 55 mph</em>
<em>Good luck !</em>
The rate of change of a linear equation (first degree) is equivalent to the slope of a line. Slope is described as the vertical movement (rise) of the line over its horizontal counterpart (run). In determining the rate of change or slope (m) given 1 data point (x',y'), point-slope form is applicable. Point-slope form is: (y-y') = m (x-x'). Substitute the given point (-5,-1) in the equation. By substitution, [y-(-1)] = m [x-(-5)]. Re-arranging the equation, the rate of change or slope is, m = (y+1)/(x+5).
Answer:
13:117 = 1:3
5:125 = 1:5
8:128 = 1:4
12:48 = 1:2
Step-by-step explanation:
117/13 = 9
√9 = 3
125/5 = 25
√25 = 5
128/8 = 16
√16 = 4
48/12 = 4
√4 = 2
Answer:
1/-5 should be the answer
