math question-Answer the given question?

Question

The elec. resistance R of a certain resistance is a function of the temp T given by the equation R=aT + b where a and be are constants. If R=1200 Ohms when T=10.0Degrees Celsius and R=1280 Ohms when T=50degrees Celsius we can find constants a and b by substituting and obtaining the equations 1200=10.0a +b; 1280=50.0a + b. Are the constants a=4.00 Ohms/Degrees Celsius and b =1160 Ohms>?

Answer 1

R = aT + b
1200 = 10a + b
1200 - 10a = b

1280 = 50a + (1200 - 10a)
1280 = 40a + 1200
80 = 40a
2 =a


1200 = 10(2) + b
1200 = 20 + b
1180 = b

Answer 2

No! Those are not correct. Set up a system of linear equations, and subtract the first equation from the second. b drops out easily. Then you can find a.

When you do this, you get:

80 = 40a.

So a obviously is 2. Substitute 2 back into either or both equations to find b. They should work out to be the same in either case. In this case, b = 1180.

a = 2 Ohms/degrees C
b = 1180 Ohms.

Substitute both calculated values into each equation to verify that they satisfy them, and you are done.

Answer 3

replace a and b by their supposed numerical values in 10.0a+b and 50.1a+b and check if you find the 1200 and 1280

10.0*4.00+1160 is indeed 1200
but 50.0*4.00+1160=1360 and not 1280

thus 4.00 and 1160 are not the constants you are looking for
(a=2.00 b=1180 are the good constants)

Answer 4

R=aT + b
substituting values:

1200=10a + b..............(i)
1280=50a + b..............(ii)

Subtracting (i) from (ii)

=>80=40a
=>a=2 ohm/deg. Celcius

=>b=1280 - 50*2=1180 ohms

Therefore, the given values are incorrect!