How many times do 12 go into 209?

Mathematics

1 answer:

Olenka [21]3 years ago

3 0

209/12 is 17.41, so 12 goes into 209 17 times.

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Will someone help with this ! I was allowed a retry for it and I don’t know how to do it

pashok25 [27]

Answer:

$\frac{dN}{dt}=k(725-N)\\separating the variables and integrating\\\int \frac{dN}{725-N}=\int kdt+c\\-log (725-N)=kt+c\\log (725-N)=-kt-c\\(725-N)=e^{-kt-c} =e^{-kt} *e^{-c} =Ce^{-kt} \\when t=0,N=400\\725-400=Ce^{0} =C\\C=325\\when t=3,N=650\\725-650=325e^{-3k} \\\frac{75}{325}=(e^{-k})^3\\\\e^{-k} =(\frac{3}{13}) ^{\frac{1}{3} } \\when t=5\\725-N=325(\frac{3}{13}) ^{\frac{5}{3} } =325*\frac{3}{13}*(\frac{3}{13} )^{\frac{2}{3}}\\N=725-75*(\frac{3}{13} )^{\frac{2}{3} }=725-28=697$

Step-by-step explanation:

6 0

3 years ago

the position of a submarine relative to the water surface was -32 1/4 feet . A downward navigational manuever increased it's dep

tester [92]

Answer:

Position of the submarine now relative to the water surface $=-47\ \frac{3}{4}$

Step-by-step explanation:

It says that the position of a submarine relative to the water surface was -32 1/4 feet . A downward navigational maneuver increased it's depth by 15 1/2 feet.

Now we need to find about what is the position of the submarine now relative to the water surface.

Downward navigation means we 15 1/2 is negative

so we will just add both values to get the final answer:

Position of the submarine now relative to the water surface = $-32\ \frac{1}{4}-15\ \frac{1}{2}$