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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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MertsjBest ResponseYou've already chosen the best response.1
Can you write the joint variation equation for x y and z? Use k for the constant.
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
Now plug in the given values and find k.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
does Mertsj's response make sense?
 one year ago

Katieann16Best ResponseYou've already chosen the best response.0
a little, im still a little confused ..
 one year ago

Katieann16Best ResponseYou've already chosen the best response.0
so it would be 32m= k *5m*12m
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
y = 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)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
32 = k*6*8 32 = k*48 k = ??
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
no it's not 4 divide both sides by 48 to solve for k
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
more like 0.667 but I would keep it as a fraction so k = 2/3
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
y = kxz y = (2/3)xz y = (2/3)*5*12 plug in x = 5 and z = 12 y = ???
 one year ago

Katieann16Best ResponseYou've already chosen the best response.0
my choices are 90m, 4m, 40m, 9m
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
now multiply that by 2 then divide that result by 3
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
close, but not quite
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
you're welcome
 one year ago
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