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# Module 4

### Variation and Problem Solving

Question | Answer |
---|---|

Direct Variation Equation | y=kx |

Inverse Variation Equation | y=k/x |

Direct Variation: y=12 x=4 | y=kx ; 12=4k ; k=(12/4)=3 ; y=3x |

Direct Variation y=4 x=6 | y=kx ; 4=6k ; k=(4/6)=2/3 ; y=(2/3)x |

Direct Variation y=2 x=1/2 | y=kx ; 2=(1/2)k ; k=(2/(1/2))=4 ; y=4x |

Inverse Variation y=10 x=15 | y=k/x ; 10=k/15 ; k=(10*15)=150 ; y=150/x |

Inverse Variation y=2 x=12 | y=k/x ; 2=k/12 ; k=(2*12)=24 ; y=24/x |

Inverse Variation y=1/5 x=1/2 | y=k/x ; 1/5=k/(1/2) ; k=(1/5*1/2)=1/10 ; y=(1/10)/x |

Joint Variation | y=kxz |

a product weighs (y) 100 lbs when it is 8' high (h) by 2'wide (w) . If the weight varies jointly with height and width, what will a 10' high8 by 10' wide piece weigh? | y=khw ; 100=k(8)(2) ; 100=16k ; k=6.25 ; y=(6.25)(10)(10) ; y=625 |

y varies jointly and directly with x and Y the square of w and inversely of z. y=100 when x=10,w=5 and z=12. find y when x=2,w=9 and z=10? | y=(kxw^2)/z ; 100=(k(10)(5^2))/12 ; k=4.8 ; y=((4.8)(2)(9^2))/10 ; y=77.76 |

Y is directly proportional to the square of X: y=2 x=8 | Y=kx^2 ; 2=k(8)^2 ; k=1/32 ; Y=(1/32)x^2 |

Y is inversely proportional to the cube of X : Y=3 X=5 | Y=k/x^3 ; 3=k/(5)^3 ; k=375 ; Y=375/x^3 |