sugargenius
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Someone pls answer these questions as quick as possible.
WARNING ⚠️ = No Silly Answers. Don’t answer if you’re not sure too.
THANKS

Someone pls answer these questions as quick as possible WARNING No Silly Answers Dont answer if youre not sure too THANKS class=

Respuesta :

DeanR

Let's do the unit cube and worry about the side length later.

Let's assign F(0,0,0), H(1,1,0), A(0,1,1), C(1,0,1)

Let's call the beetle positions P and Q, functions of time t.

P = F + 2t(H-F) = tH = 2t(1,1,0) = (2t,2t,0)

Q = A + t(C-A) = (0,1,1)+t(1,-1,0)=(t, 1-t, 1)

The squared distance is

[tex]PQ^2 = (2t - t)^2 + (2t -(1-t))^2 + 1 = t^2 + (3t-1)^2 + 1 = 10t^2-6t+2[/tex]

[tex]\dfrac{d\ PQ^2}{dt} = 20t - 6 = 0[/tex]

[tex]t = 6/20 = 3/10 [/tex]

[tex]PQ^2 = 10(3/10)^2 - 6(3/10) + 2 = 11/10[/tex]

[tex]\sqrt{11/10} \times 40 \sqrt{110} = 440 [/tex]

Answer: 440

Let's do the unit cube and worry about the side length later.

Let's assign F(0,0,0), H(1,1,0), A(0,1,1), C(1,0,1)

Let's call the beetle positions P and Q, functions of time t.

P = F + 2t(H-F) = tH = 2t(1,1,0) = (2t,2t,0)

Q = A + t(C-A) = (0,1,1)+t(1,-1,0)=(t, 1-t, 1)

The squared distance is

PQ^2 = (2t - t)^2 + (2t -(1-t))^2 + 1 = t^2 +  (3t-1)^2 + 1 = 10t^2-6t+2

\dfrac{d\  PQ^2}{dt} = 20t - 6 = 0

t = 6/20 = 3/10

PQ^2 = 10(3/10)^2 - 6(3/10) + 2 = 11/10

\sqrt{11/10} \times 40 \sqrt{110} = 440

Answer: 440

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