Answer:
91.6 km/h
Step-by-step explanation:
Let v = average speed of van = 65 km/h and v' = average speed of car. Let t be the time the car starts to move. The van started 35 minutes earlier at t' = (t + 35/60) h.
Since distance d = vt where v = velocity and t = time, the distance moved by the van d = vt' = v(t + 35/60) and that moved by the car is d' = v't.
Since the car catches up with the van after it had moved a distance of 130 km, d = d' = 130 km.
So d = v(t + 0.583)
Substituting d = 130 km and v = 65 km/h, we have
130 km = 65 km/h(t + 0.583)
130 km = (65t + 37.895 )km
subtracting 37.895 from both sides, we have
130 km - 37.895 km = 65t
92.105 = 65t
dividing both sides by 65, we have
t = 92.105/65
= 1.417 h
≅ 1.42 h
Since d = d' = v't,
v' = d'/t
= 130 km/1.42 h
= 91.55 km/h
≅ 91.6 km/h
So, the average speed of the car is 91.6 km/h
Answer:
B
Step-by-step explanation:
Range of the graph is the ALLOWED y-values. The y-axis is number of gallons left in tank. So, <u><em>it cannot be NEGATIVE number of gallons, so 0 is the lower limit of the range.</em></u>
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As we can see from the axis of the graph, we see where the line cuts the y-axis, that is the upper limit of number of gallons he starts off with. The y-intercept (y-axis cutting point) is 12.
So we can say that the range is 0 ≤ y ≤ 12
Correct answer is B
9514 1404 393
Answer:
BD = 17
Step-by-step explanation:
The Pythagorean theorem can be used twice:
BC^2 + AC^2 = AB^2
BC = √(AB^2 -AC^2) = √(25^2 -24^2) = √49 = 7
__
Likewise, ...
DC = √(AD^2 -AC^2) = √(26^2 -24^2) = √100 = 10
Then ...
BD = BC +DC = 7 + 10
BD = 17
The answer would be 317.03
Ricardo is not correct because the temperature is actually getting colder (idk if I need to explain it more). The difference is -12 F
