Juliet and mercutio are moving at constant speeds in the xy-plane. they start moving at the same time. juliet starts at the point (0, −6) and heads in a straight line toward the point (10, 5), reaching it in 10 seconds. mercutio starts at (7, −16) and moves in a straight line. mercutio passes through the same point on the x-axis as juliet, but 2 seconds after she does. how long does it take mercutio to reach the y-axis?

  574/17 ≈ 33.765 seconds

Step-by-step explanation:

This problem begs to be solved using parametric equations for the lines of travel. Such an equation for position (P) can be written in terms of a starting point (A) and direction of travel (V) as ...

  P = A + tV

Here, the direction of travel between points A and B is V = B-A, and the time value (t) is scaled by the amount of time it takes to get there.

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The position of Juliet as a function of time from point A to point B in 10 seconds can be written as ...

  J = A + (t/10)(B-A) = (0, -6) + (t/10)(10-0, 5-(-6))

  J = (t, 1.1t -6)

Juliet will reach the x-axis when her y-position is zero:

  1.1t -6 = 0

  t = 6/1.1 = 60/11

Her position at that point is J = (60/11, 0).

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Mercutio's position can be described similarly. Starting at point C(7, -16), he reaches D(60/11, 0) after 60/11+2 = 82/11 seconds.

  M = C +(11t/82)(D -C) = (7, -16) +(11t/82)(60/11 -7, 0 -(-16))

  M = (7 -17/82t, 88/41t -16)

Mercutio will reach the y-axis when his x-position is zero:

  7 -17/82t = 0

  7 = 17/82t

  82/17×7 = t = 574/17 ≈ 33.765 . . . . seconds

It takes Mercutio about 33.765 seconds to reach the y-axis.

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6

a

step-by-step explanation: