PLEASE HELP- John and Ariana bought school supplies. John spent $10.60 on 5 notebooks and 5 pens. Ariana spent $7.30 on 3 notebo
1 answer:
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6x+42°=180°(ANGLES ON A STRAIGHT LINE)
IF DF IS PERPENDICULAR TO AC IT MEANS m<DBC =90°
IF DF IS PERPENDICULAR TO AC IT MEANS m<DBC =90°m<DBE+m<EBC= m<DBC
IF DF IS PERPENDICULAR TO AC IT MEANS m<DBC =90°m<DBE+m<EBC= m<DBCm<DBE+18°=90°
IF DF IS PERPENDICULAR TO AC IT MEANS m<DBC =90°m<DBE+m<EBC= m<DBCm<DBE+18°=90°m<DBE =90°-18°
IF DF IS PERPENDICULAR TO AC IT MEANS m<DBC =90°m<DBE+m<EBC= m<DBCm<DBE+18°=90°m<DBE =90°-18°mDBE=72°
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Answer:
(c) dZ/dt = 11/√544, so moving away at about 0.47 m/s
Step-by-step explanation:
The (x, y) coordinates of the car are related by the fact that they are on a circle centered at (12, 7) with a radius of 5. Then their rates of change are related by the derivative of the circle equation with respect to time.
(x -12)^2 + (y -7)^2 = 25
2(x -12)x' +2(y -7)y' = 0
y' = -(x -12)/(y -7)x'
At the time and point of interest, we have ...
y' = -(15 -12)/(11 -7)(1) = -3/4
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The distance (z) to the observer is given by the Pythagorean theorem:
z^2 = (x -0)^2 +(y -2)^2
and its rate of change with time is ...
2z·z' = 2(x-0)x' +2(y -2)y'
The distance d at the point of interest is ...
z = √((15 -0)^2 +(11 -2)^2) = √(225 +81) = √306 = 3√34
So, the rate of change of distance to the observer at the time and point of interest is ...
z' = (x(x') +(y -2)y')/z
z' = ((15)(1) +(11-2)(-3/4))/(3√34) = (33/4)/(3√34)
z' = 11/√544 ≈ 0.47 . . . m/s
