66.2k views
0 votes
If (x) varies directly as (y) and (x = 150) when (y = 5), find:

a) (xy = 750)
b) (x = 30y)
c) (x + y = 155)
d) (x/y = 30)

User Eliasdx
by
8.2k points

1 Answer

3 votes

Final answer:

In direct variation where x = 30y, solutions for the conditions xy = 750, x + y = 155, and ratios x = 30y or x/y = 30 affirm the direct relationship, yielding x = 150 and y = 5 whenever these conditions are applied.

Step-by-step explanation:

When x varies directly as y, there exists a constant k such that x = ky. Given that x = 150 when y = 5, we find the constant using the equation 150 = k × 5. Solving for k, we get k = 30. This constant allows us to solve for other relations involving x and y.

  • a) To find x when the product xy = 750, we substitute k into the direct variation equation x = 30y and set it equal to 750: 30y × y = 750. Solving for y, we find that y = 5 and x = 150.
  • b) When x = 30y, this is simply the direct variation equation we've already established, meaning this relationship is always true for any values of x and y that maintain the variation.
  • c) For x + y = 155, substitute x = 30y into the equation: 30y + y = 155. This simplifies to 31y = 155. Dividing both sides by 31, we find that y = 5 and therefore x = 150.
  • d) The ratio x/y = 30 reflects the constant of variation; thus, this equation is also always true for any set of values x and y that maintain the direct variation.

User Gabe
by
8.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories