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# Variation Flash Card

### Variation and Problem Solving

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

Y varies directly as X, if Y is 6 when X is 36, find the constant of variation and the direct variation equation. | Y=KX.Y=6 X=36 .Substitute values. 6=K(36). Solve, K=6/36.Simplify, K=1/6. Rewrite with K value for the direct variation equation. Y=1/6X |

Hooke's Law states that the distance a spring stretches is directly proportional to the weight attached to a spring. If a 50 pound weight is attached to a spring that stretches 10 inches, how much would an 80 pound weight stretch it? | D=KW D= distance stretched, c.o.v.= K, W= weight attached.Substitute values, 10=K(50). Solve, K=10/50, K=1/5. Now solve for the distance stretched by using K. D=1/5(80). The spring would stretch 16 inches. |

Suppose the letter V varies inversely with W. If V is 5 when W is 4 find the constant of variation and inverse variation equation. | V=K/W. V=5, W=4.Substitute values into equation,5=K/4. Solve, K=20. Write inverse variation equation, V=20/W. |

Boyle's Law states that if temp stays the same, the pressure P of gas is inversely proportionate to the Volume V. If a cylinder has a pressure of 810 kilo pascals when the volume is 1.5 cubic meters, find the pressure when volume is 2.5 cubic meters. | P=K/V. P=810,V=1.5.Substitute values, 810=K/1.5 .Solve,K=810(1.5), K=1215. P= 1215/2.5, so P=486. The pressure is 486 kilo pascals. |

If Y varies jointly as X and Z, and Y=12 when X=9 and Z=3, find z when y=6 and x=15. | Y=KXZ. Y=12,X=9,Z=3.Substitute values, 12=K(9)(3). Solve, 12=K(27),K=12/27, K=4/9.Now substitute K value in to solve for Z when Y=6 and X=15, 6=(4/9)(15)(Z), 6=(60/9)Z,54=60Z,Z=9/10. |

Suppose Y varies directly with the square of X. If Y is 36 when X is 2, find the constant of variation and the direct variation equation. | Y=KX^2. Y=36, X=2.Substitute values, 36=K(2)^2. Solve, 36=K(4),36/4=K, K=9.The equation is, Y=9X^2. |

If Y varies directly as X and inversely as Z, and Y = 24 when X=48 and Z=4, find X when Y=44 and Z = 6. | Y=KX/Z. Y=24,X=48,Z=4. Substitute values, 24=K(48)/4. Solve, 24=12K,K=2.Put in value of K to find X when Y=44. Y=2X/Z,44=2X/6,44=X/3,X=132. |

If Y varies directly as X and inversely as Z squared, and Y= 12,X=1, and Z=2, find X when Y=16, and Z=4.Round to nearest decimal point. | Y=KX/Z^2.Y=12,X=1,X=2.Substitute Values, 12=K(1)/2^2. Solve, 12=K/4,K=48.Now substitute K value in to solve for X when Y=16 and Z=4. 16=48(X)/4^2, 16=48X/16, 16=3X, X=5.3 |

If Y varies directly with the square of X, and Y= 32 when X=2, find X when Y=56. Round to nearest decimal point. | Y=KX^2.Y=32,X=2.Substitute values, 32=K(2^2),32=4K,32/4=K,K=8.Now substitute Y and K values to solve for X. 56=8X^2,56/8=X^2,7=X^2,The square root of 7=X, X=2.6. |

If R varies directly with L,and inversely with M,when R= 30 and L=3 and M=2. Find L when R=90 and M=3. | R=KL/M.R=30,L=3,M=2.Substitute values, 30=K(3)/2. Solve, 30=K(3)/2, 60=3K, K=20. Now put in K's value to solve for L when R=90 and M=3. 90=20L/3, 270=20L,270/20=L, L=13.5. |

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