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Sonja [21]
3 years ago
14

Please Answer Please answer ASAP illg ive brainliest

Mathematics
1 answer:
Luba_88 [7]3 years ago
4 0

Answer:

Speed = 100/Time

Step-by-step explanation:

(a). Speed is a function of time : As seen from equation (1) Speed of the students varies according to the value of time. So, This a true statement.

(b). Time is a function of distance : Distance is not varying in the equation (1), it is fixed to be 100. So, time is not a function of distance because here the distance is made constant. So, This is false statement.

(c). Speed is a function of number of students racing : Speed is not depending on the number of students racing so sped is independent of the number of the students racing. Therefore, Speed is not a function of number of students racing. This is False statement.

(d). Time is a function of speed : As seen from equation (1) Time of the students varies according to the value of their speed. So, This a true statement.

Hence, the true statements are (a) and (d)

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I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

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L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

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\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
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