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Zigmanuir [339]
4 years ago
14

dale is driving to Miami. suppose that the distance (in miles) is a linear function of his total driving time (in minutes). dale

has 59 miles to his destination after 40 minutes of driving, and he had 50 miles to his distination after 52 minutes of driving. how many miles will he have to his destination after 70 minutes of driving?
Mathematics
1 answer:
nexus9112 [7]4 years ago
3 0
Considering there is a function (relationship) and that it is linear, the distance will change proportionally to time constantly. In other words, we are taking the speed to be constant throughout the journey.
If we let:
t = time (min's) driving
d = distance (miles) from destination
Then we can represent the above information as:
t = 40: d = 59
t = 52: d = 50
If we think of this as a graph, we can think of the x-axis representing time and the y-axis representing the distance to the destination. Being linear, the function will be a line, i.e. it will have a constant gradient. If you were plot the two points inferred from the information and connect the two dots, you will get a declining line (one with a negative gradient) representing the inversely proportional relationship or equally, the negative correlation between the time driving and the distance to the destination. The equation of this line will be the linear function that relates time and the distance to the destination. To find this linear function, we do as follows:

Find the gradient (m) of the line:
m = Δy/Δx
In this case, the x-values are t-values and our y-values are d-values, so:
Δy = Δd
= 50 - 59
= -9
Δx = Δt
= 52 - 40
= 12
m = -9/12 = -3/4
Note: m is equivalent to speed with units: d/t

Use formula to find function and rearrange to give it in the desired format:
y - y₁ = m(x - x₁)
d - 50 = -3/4(t - 52)
4d - 200 = -3t + 156
4d + 3t - 356 = 0

Let t = 70 to find d at the time:
4d + 3(70) - 356 = 0
4d + 210 - 356 = 0
4d - 146 = 0
4d = 146
d = 73/2 = 36.5 miles

So after 70 min's of driving, Dale will be 36.5 miles from his destination.
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<em>Additional comments</em>

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