Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

What does inversely proportional mean?

OpenStudy (anonymous):

Inversely Proportional: when one value decreases at the same rate that the other increases.

OpenStudy (anonymous):

examples?

Nnesha (nnesha):

i don't think s i can explain but i know the equation for inverse prp \[\huge\rm y=\frac{ k }{ x }\]haha :P

OpenStudy (anonymous):

What is the K?

OpenStudy (zale101):

K is constant

Nnesha (nnesha):

^what she said :) constant of variation

OpenStudy (zale101):

In this equation, we know that if y increases, x decreases and vice versa

OpenStudy (anonymous):

The time it takes to erect a bonfire is inversely proportional to the number of students doing the work. If it takes 20 students 1.5 hours to do the job, how long will it take 35 students to do the job, to the nearest minute?

Nnesha (nnesha):

time = y students= x

OpenStudy (anonymous):

THis is a super confusing problem

Nnesha (nnesha):

okay equation for inverse variation is y =k/x <--(y inversely proportional to x) read the question **time it takes to erect a bonfire is inversely proportional to the number of students** time inversely proportional to number of students let time = t students = S so equation would be \[\huge\rm t =\frac{ k }{ s }\]

Nnesha (nnesha):

when t =1.5 s=20 so plug in t and s value solve for k (constant of variation)

OpenStudy (anonymous):

ohhhh okay so 30

Nnesha (nnesha):

that's k you need y value when =35 so now use the same equation solve for y

Nnesha (nnesha):

\[\rm y=\frac{ 30 }{ 35 }\]

OpenStudy (anonymous):

6/7

OpenStudy (anonymous):

6/7 x 60 = 360/7 = 51.43 minutes

Nnesha (nnesha):

***to the nearest minute*** so 60 sec in a mint multiply that by 60

OpenStudy (anonymous):

so 51 minutes?

Nnesha (nnesha):

ye

OpenStudy (anonymous):

got it thanks!!

Nnesha (nnesha):

my pleasure

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!