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Katieann16
 3 years ago
y varies jointly as x and z, and y = 32 m, x = 6 m, and z = 8 m.
What is the value of y when x = 5 m and z = 12 m?
Katieann16
 3 years ago
y varies jointly as x and z, and y = 32 m, x = 6 m, and z = 8 m. What is the value of y when x = 5 m and z = 12 m?

This Question is Closed

Mertsj
 3 years ago
Best ResponseYou've already chosen the best response.1Can you write the joint variation equation for x y and z? Use k for the constant.

Mertsj
 3 years ago
Best ResponseYou've already chosen the best response.1Now plug in the given values and find k.

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1does Mertsj's response make sense?

Katieann16
 3 years ago
Best ResponseYou've already chosen the best response.0a little, im still a little confused ..

Katieann16
 3 years ago
Best ResponseYou've already chosen the best response.0so it would be 32m= k *5m*12m

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1y = kxz we're given y = 32 m, x = 6 m, and z = 8 m plug all this in to get 32 = k*6*8 now solve for k Note: I think we can ignore the m's because I think they refer to meters (eg: x = 6 m means x is 6 meters long)

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.132 = k*6*8 32 = k*48 k = ??

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1no it's not 4 divide both sides by 48 to solve for k

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1more like 0.667 but I would keep it as a fraction so k = 2/3

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1y = kxz y = (2/3)xz y = (2/3)*5*12 plug in x = 5 and z = 12 y = ???

Katieann16
 3 years ago
Best ResponseYou've already chosen the best response.0my choices are 90m, 4m, 40m, 9m

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1now multiply that by 2 then divide that result by 3

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1close, but not quite

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1you're welcome
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