Questions: 36. x is directly proportional to the cube of y. x=3 when y=7. What is x when y=14? x= 1. 24 2. 196 3. 12 4. 21 37. 8x+27+3y=27y+3; x= 1. 13y+3 2. 3y-3 3. 3y 4. 6

36. x is directly proportional to the cube of y. x=3 when y=7. What is x when y=14? x=
1. 24
2. 196
3. 12
4. 21

37. 8x+27+3y=27y+3; x=
1. 13y+3
2. 3y-3
3. 3y
4. 6
Transcript text: 36. x is directly proportional to the cube of y . $\mathrm{x}=3$ when $\mathrm{y}=7$. What is x when $\mathrm{y}=14$ ? $\mathrm{x}=$ 1 24 2 196 3 12 4 21 37. $8 x+27+3 y=27 y+3 ; \quad x=$ 1 $13 y+3$ 2 $3 y-3$ 3 $3 y$ 4 6
failed

Solution

failed
failed

Solution Steps

Solution Approach
  1. For the equation \(8 + 7 = 5x - 15\), solve for \(x\) by isolating it on one side of the equation.
  2. For the problem where \(x\) is directly proportional to the cube of \(y\), use the proportionality relationship to find the constant of proportionality and then calculate \(x\) for \(y = 14\).
  3. For the equation \(8x + 27 + 3y = 27y + 3\), rearrange the terms to solve for \(x\) in terms of \(y\).
Step 1: Solve the Equation \(8 + 7 = 5x - 15\)

Starting with the equation: \[ 8 + 7 = 5x - 15 \] we simplify it to: \[ 15 = 5x - 15 \] Adding 15 to both sides gives: \[ 30 = 5x \] Dividing both sides by 5 results in: \[ x = 6 \]

Step 2: Find \(x\) When \(y = 14\) Given \(x\) is Proportional to \(y^3\)

Given that \(x\) is directly proportional to the cube of \(y\), we can express this as: \[ x = k y^3 \] where \(k\) is the constant of proportionality. We know that \(x = 3\) when \(y = 7\): \[ 3 = k (7^3) \implies k = \frac{3}{343} \approx 0.008746355685131196 \] Now, substituting \(y = 14\): \[ x = k (14^3) = 0.008746355685131196 \times 2744 = 24.0 \]

Step 3: Solve the Equation \(8x + 27 + 3y = 27y + 3\)

Starting with the equation: \[ 8x + 27 + 3y = 27y + 3 \] Rearranging gives: \[ 8x = 27y - 3y + 3 - 27 \] which simplifies to: \[ 8x = 24y - 24 \] Dividing both sides by 8 results in: \[ x = 3y - 3 \]

Final Answer

The solutions are:

  1. \(x = 6\)
  2. \(x = 24\) when \(y = 14\)
  3. \(x = 3y - 3\)

Thus, the final boxed answers are: \[ \boxed{x = 6} \] \[ \boxed{x = 24} \] \[ \boxed{x = 3y - 3} \]

Was this solution helpful?
failed
Unhelpful
failed
Helpful