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Alja [10]
2 years ago
8

Can anyone help me with this please .

Mathematics
1 answer:
Mazyrski [523]2 years ago
8 0

Answer:

Left: yes it is linear

Right: No it is not.

Step-by-step explanation:

Left Problem

The table on the left is linear. It's general form is

y = 6x + 7

x = 5

y = 5*6 + 7

y = 37

-------------------

x = 8

y = 6*8 + 7

y = 48 + 7

y = 55

------------------

y = 6*12 + 7

y = 72 + 7

y = 79

=============

Right Problem

The right problem looks like it is linear. It is not, I don't think. Do the same thing you did above.

y = 62x

x = 2

y = 2*62

y = 124

-----------------------

y = 62x

x = 6

y = 62 * 6

y = 372

---------------------

y = 62*x

x = 9

y = 62 * 9

y = 558            

It should be 558 not 552

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Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te
Aleonysh [2.5K]

Answer:

t=5.18 minute

Step-by-step explanation:

Using Newton's law of cooling

\frac{dT}{dt}=-k(T-T_{s})

T is temperature of a cup of coffee at any time t T_{s}, is the temperature of the surrounding and k is a constant of proportionality in this negative because temperature is decreasing.

T(0)=190\\T(1)=170\\T_{s}=66

\frac{dT}{dt}=-k(T-T_{s})\\\frac{dT}{(T-T_{s})}=-k*dt\\\int\limits {\frac{1}{T-T_{s} } } \, dT=-\int\limits {k} \, dt\\ln(T-T_{s})=k*t+C\\e*ln(T-T_{s})=e^{k*t+C} \\T-T_{s}=e^{k*t}*e^{C}\\e^{C}=C\\T-T_{s}=e^{k*t}*C\\T=T_{s}+e^{k*t}*C

To find constant knowing the time and the temperature the first step of the change of energy in cup of coffee

T=T_{s} +e^{-k*t}*C\\ C=190-66=124\\170=66+124*e^{-k*2}\\ 170-66=124*e^{-k*2}\\ln(\frac{104}{124})=ln*e^{-k*2}\\-0.175=-k*2\\k=0.08794

Now using the constant of decreasing can find the time to be a 145 temperature the cup of coffee

145=66+124*e^{-k*t} \\145-66=124*^{-0.087*t}

ln(0.637)=ln*e(-0.087*t)\\-0.45=-0.087*t\\t=5.18minutes

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Step-by-step explanation:

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