Question

Can anyone explain how I can put them together? I can work out the individual proportions and their equations I just can’t remember how to put them together to find C.

Answer 1

Hi,
Answer is the 5th formula

Explanation:

`A\ is\ inversely\ proportional\ to\ B\ squared:\\\\A=(k_1)/(B^2) \\if\ B=4\ then\ A=(1)/(8) =(k_1)/(16) \\\Longrightarrow\ k_1=(16)/(8)=2\\\\B^2=(2)/(A) \Longrightarrow\ B=\sqrt{(2)/(A) } \\\\`

`C\ is\ directly\ proportional\ to \ B^3\\\\C=k_2*B^3\\if\ B=5\ then\ C=50: \\50=k_2*5^3\ \Longrightarrow\ k_2=(50)/(125) =(2)/(5) \\\\\\C=(2)/(5) *B^3=(2)/(5) *(\sqrt{(2)/(A) })^3=(4)/(5)*(√(2) )/(A^(1.5))`

Answer 2

E

Explanation:

given A is inversely proportional to B² then the equation relating them is

A =
`(k)/(B^2)` ← k is the constant of variation

to find k substitute B = 4 , A = 0.125 into the equation

0.125 =
`(k)/(4^2)` =
`(k)/(16)` ( multiply both sides by 16 )

2 = k

A =
`(2)/(B^2)` ← equation of variation → (1)

given C is directly proportional to B³ then the equation relating them is

C = kB³

to find k substitute B = 5, C = 50 into the equation

50 = k × 5³ = 125k ( divide both sides by 125 )

`(50)/(125)` = k , that is

k =
`(2)/(5)`

C =
`(2)/(5)` B³ → (2)

to find B³ in terms of A use (1)

A =
`(2)/(B^2)` ( multiply both sides by B²

AB² = 2 ( divide both sides by A )

B² =
`(2)/(A)` ( take square root of both sides )

B =
`\sqrt{(2)/(A) }`

then

B³ =
`\sqrt{(2)/(A) }` ×
`\sqrt{(2)/(A) }` ×
`\sqrt{(2)/(A) }` =
`(2)/(A)` ×
`\sqrt{(2)/(A) }` =
`(2√(2) )/(A√(A) )` substitute into (2)

C =
`(2)/(5)` ×
`(2√(2) )/(A√(A) )` =
`(4√(2) )/(5A^(1.5) )`