It costs a family $324.00 to buy a ten-ft-by-12-ft rug. If the cost is proportional to the area, what will it cost the family to buy a 15-ft-by-18-ft rug?
Do you know what it means for the cost to be proportional to the area?
no
This means there is some constant, let's call it, k such that cost=k*area
ok now we have an equation do you know how to find the area of a 10 ft by 12 ft rug?
multiply them?
right
so the area=10*12 ft^2 when we have the cost is 324 this is a point on our line cost=k*area use this point (area=10*12, cost=324) to find k
solve for k and I will check your k then we will move on to the final question of this problem
I am very confused
we were given cost=k*area when it told us that the cost was proportional to the area we were also given when the area=10*12 we have the cost=324 this is enough information to find our constant of proportionally (aka k)
you need to enter area=10*12 and cost=324 into cost=k*area to find k do you still not understand what I'm asking you to do? please let me know.
324=k*120
right solve for k
324/120=2.7
cool so you just found the constant of proportionality
so we can find the cost for any area rug or find the area of the rug given the cost
so we have cost=2.7*area
what will it cost the family to buy a 15-ft-by-18-ft rug this is your question
you are given the area basically and you are asked to find the cost using the equation cost=2.7*area
cost=2.7*270 cCost=$729.00
beautiful
you want me to show a slightly different way you could have done this problem ?
you could have just had one equation instead of even bothering with the constant of proportionality that is you could have just solved \[\frac{\text{ area of second rug }}{\text{ area of first rug }}=\frac{\text{ cost of second rug }}{\text{ cost of first rug }} \\ \frac{x}{324}=\frac{270}{120} \\ \text{ multiply both sides by } 324 \\ x=\frac{270}{120} \cdot 324=729\]
if you didn't like the first way please read that last post
lol I put the costs on the opposite side but whatever a=b means b=a :p
Join our real-time social learning platform and learn together with your friends!