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jiggyN
22.12.2020 •
Mathematics
Suppose a blue car is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph (miles per hour). At the same time, a red car is 1/2 mile east of the intersection, driving east (away from the intersection) at an unknown speed. The the straight-line distance between the blue and red car is increasing at a rate of 15 mph. What is the speed of the red car?
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Ответ:
dx/dt = - 18,79 mph
Step-by-step explanation:
The two cars with the intersection point and the straight-line distance between the cars make up a right triangle. In that right triangle, the legs are the distance between each car and the intersection point, and the distance between cars is the hypothenuse
If we call x and y distances between blue car and red car respectively and L the hypothenuse by Pythagoras theorem we have:
L² = x² + y² (1)
Tacking derivatives on both sides of the equation
2*L*dL/dt = 2*x*dx/dt + 2*y*dy/dt
And from equation (1)
L² = (0,5)² + (0,5)² ⇒ L = √(0,5)² + (0,5)² ⇒ L = 0,5*√2
By subtitution in equation (2)
2*(0,5*√2)*15 = 2*0,5*dx/dt + 2*0,5*40
(15*√2 - 40 ) / 1 = dx/dt [mph]
dx/dt = - 18,79 mph
Note the( - ) sign is equivalent to say that the car is driving away from the intersection point
Ответ:
A line segment from a vertex to the midpoint of the opposite side is a "median". A median divides the area of the triangle in half, as it divides the base in half without changing the altitude.
AAMC is half AABC. AADC is half AAMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
ABMC is half AABC. ABMD is half ABMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
Then, AADC = 1/4 AABC = ABMC, so AADC = ABMC by the transitive property of equality.
Step-by-step explanation: