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20 June, 00:40

If y varies jointly as x and the cube of z and y=16 when x=4 and z=2 then y=0.5 when x=-8 and z=-3

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Answers (2)
  1. 20 June, 00:48
    0
    y = 1/2xz³ and y = 108 when = - 8 and z = - 3

    Step-by-step explanation:

    From question statement, we observe that

    y ∝ xz³

    y = kxz³ eq (1)

    Where k is proportionality constant.

    Given that

    y = 16 when x = 4 and z = 2

    k = ?

    Putting the given values in above eq (1), we have

    16 = k (4) (8)

    16 = 32k

    k = 16/32

    k = 1/2

    putting the value of k in eq (1), we have

    y = 1/2xz³ eq (2)

    Putting x = - 8 and z = - 3 in eq (2), we have

    y = 1/2 (-8) (-3) ³

    y = 1/2 (-8) (-27)

    y = 1/2 (216)

    y = 108 when x = - 8 and z = - 3.
  2. 20 June, 01:06
    0
    k = 1/2

    y = - 108

    Step-by-step explanation:

    y varies jointly as x and the cube of z

    y = kxz^3

    We know y=16 when x=4 and z=2 so we can determine k

    16 = k (4) 2^3

    16 = k*4*8

    16 = 32k

    Divide each side by 32

    16/32 = 32k/32

    1/2 = k

    y = 1/2 xz^3

    Now y=? when x=-8 and z=-3

    y = 1/2 * 8 (-3) ^3

    y = (1/2) * 8 (-27)

    y = 4 (-27)

    y = - 108
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