C= k - 273.15 (a) write two other related equations . (b) if the temperature is 23 in degrees celsius , what is the temperature in kelvins? show your work. (c) which is greater, 300 kelvins or 25 degrees celsius ? show your work.

The two other related equations are:

k = C + 273.15

    2. k/C = 273.15

K =  C + 273.15

C = 23 degrees

Hence:

K = 23 + 273.15

K = 296.15 K

23 degrees celsius is 296.15 Kelvins

K =  C + 273.15

C = 25 degrees

Hence:

K = 25 + 273.15

K = 298.15 K

25 degrees celsius is 298.15 Kelvins

Therefore, 300 Kelvins is greater than 25 degrees

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(a) k = C + 273.15, F = 1.8(k - 273.15) + 32

(b) k = 296.15

(c) 25 celcius is greater than 300 kelvis.

Step-by-step explanation:

(a) one equation would be to solve for k instead of c, and another related equation could be from kelvin to Farenheit.

C = k - 273.15

C + 273.15 = k

k = C + 273.15

F = 1.8(k - 273.15) + 32

(b) for this one you have to substitute in the previous equation and solve for k.

23 = k - 273.15

k = 23 + 273.15

k = 296.15

(c) you have to convert either kelvins to celcius or celcius to kelvins.

C = k - 273.15

C = 300 - 273.15

C = 26.85

So 25 celcius is greater than 300 kelvis.

d? maybe

step-by-step explanation:

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: