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(1) y varies partly as D an partly as the cube root of D.when Y=6, D=27 an when Y=11,D=1/8.find Y when D=3. Correct to 2 decimal point.

(2) using completing the square method find the root of the equation
8x+6x-5=0

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2 answers
  1. Y=k(D)+j(D^1/3)

    6=27k + 3j
    11=k/8+j/2

    multiply second equation by 6, then subtract first equation from second.
    6=27k+3j
    66=3k/4+5j
    or
    60=-k(3/4-27)
    solve for k
    then put that k into either equation, solve for j.

    Now solve for y when D=3

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    bobpursley
  2. 1) y = aD + b (cube root of D)
    a and b are constant.
    when y=6, d=27
    6= 27a + b(3√ 27)
    6= 27a + 3b ----------equation 1

    when y= 11, d = 1/8
    11= a/8 + b( 3√ 1/8)
    11= a/8 + b/2 --------- equation 2

    so we have
    6= 27a + 3b
    11= a/8 + b/2
    Rearrange
    27a + 3b =6
    a/8 + b/2 =11

    Multiply equation 2 by 6 and equation 1 by 1
    27a + 3b =6
    6a/8 + 6b/2 =66

    therefore,
    27a + 3b =6
    3a/4 + 3b =66

    Eliminate 3b in both equation
    27a =6
    3a/4=66

    subtract equ 1 from 2

    -105a/4 = 60
    Cross Multiply
    -105a=240

    a= -2.28

    Solve for b
    6= 27a + 3b
    6= 27(-2.28) + 3b
    3b=6 - (-61.56)
    3b=67.56
    b=22.52

    Now solve for y when D=3
    y = aD + b (cube root of D)
    y = (-2.28 x 3) + 22.52 (cube root of 3)
    y= -6.84 + 32.48
    y= 25.64 to 2 D.P

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