Ningalairamaeidn
Answered

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SOLVING:

1. y varies directly as z and inversely as x and y = 8 when z = 4 and x = ⅘

2. y varies directly as x and inversely as z and y = 6 when x = 3 and z = 2​

Sagot :

UK2019idn

✏️COMBINED VARIATIONS

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[tex]\underline{\mathbb{PROBLEMS:}}[/tex]

  • #1. y varies directly as z and inversely as x and y = 8 when z = 4 and x = ⅘.
  • #2. y varies directly as x and inversely as z and y = 6 when x = 3 and z = 2.

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[tex]\underline{\mathbb{ANSWERS:}}[/tex]

[tex]\qquad\LARGE\rm»\:\: 1. \:\green{k=\frac58}[/tex]

[tex]\qquad\LARGE\rm»\:\: 2. \:\green{k=4}[/tex]

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[tex]\underline{\mathbb{SOLUTIONS:}}[/tex]

- Write the equation for combined variations and then find the constant k.

#1.

  • [tex]y = \frac{kz}{x} \\ [/tex]

  • [tex]8 = \frac{k(4)}{ \frac{4}{5} } \\ [/tex]

  • [tex]8 = \frac{4k}{ \frac{4}{5} } \\ [/tex]

  • [tex]8 = 4k \div \frac{4}{5} \\ [/tex]

  • [tex]8 = 4k \cdot \frac{5}{4} \\ [/tex]

  • [tex]8 = \frac{20k}{4} \\ [/tex]

  • [tex]8 = 5k[/tex]

  • [tex] \frac{8}{5} = \frac{ \cancel5k}{ \cancel5} \\ [/tex]

  • [tex] \frac{8}{5} = k \\ [/tex]

[tex]\therefore[/tex] The constant of the variation is 8/5.

[tex]\rm[/tex]

#2.

  • [tex]y = \frac{kx}{z} \\ [/tex]

  • [tex]6 = \frac{k(3)}{2} \\ [/tex]

  • [tex]6 = \frac{3k}{2} \\ [/tex]

  • [tex]6 \cdot2 = \frac{3k}{ \cancel2} \cdot \cancel2 \\ [/tex]

  • [tex]12 = 3k[/tex]

  • [tex] \frac{12}{3} = \frac{ \cancel3k}{ \cancel3} \\ [/tex]

  • [tex]4 = k[/tex]

[tex]\therefore[/tex] The constant of the variation is 4.

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#CarryOnLearning

Glennreg23idn

Answer:

1. 8/5

2. 4

Step-by-step explanation:

hopeithelps

brainliest the owner of this answer:)

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