SOLUTION: the temperature at which water boils is calculated using the formula;T=ma+b where T=temperature in degrees Celsius and a=altitude in metres. In Banff Alberta, the altitude is 1383m
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Question 101818: the temperature at which water boils is calculated using the formula;T=ma+b where T=temperature in degrees Celsius and a=altitude in metres. In Banff Alberta, the altitude is 1383m and water boils at 95 degrees Celsius.At the peak of Mt. Logan BC, the altitude is 5951m and the water boils at 80 degrees Celsius. Deteremine the values of m and b.
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Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
T=ma+b
First, state what you know from the data. Equation (1) uses the data from Banff.
(1) 95 = m(1383)+b
Equation (2) uses the data from Mt. Logan.
(2) 80 = m(5951)+b
Good news. Two equations, two unknowns, there is a solution. Use one equation to solve for b and then substitute into the other equation to solve for m.
Let’s use equation (2) and solve for b.
(2) 80 = m(5951)+b
80 = m(5951)+b
80-m(5951) = m(5951)-m(5951)+b Additive inverse of m(5951) or (-m(5951))
80-m(5951) = b or
(3) b = 80-m(5951) We’ll call this equation (3) but it’s really (2) re-arranged.
Use this in equation (1) and solve for m.
(1) 95 = m(1383)+b
95=m(1383)+(80-m(5951))
95=m(1383)-m(5951)+80 Simplify
95=-4568m+80
95-80 = -4568m+80-80 Additive inverse of 80 or (-80).
15=-4568m
15/(-4568)=-4568m/(-4568) Multiplicative inverse of -4568 or (1/-4568).
-15/4568 = m or
m = -15/4568
Substitute this value into (1), (2), or (3) to solve for b. Equation (3) is the most direct.
(3) b = 80-m(5951)
b = 80 – (-15/4568)(5951)
b = 80 + (15/4568)(5951)
b = 80 + (89265/4568)
Those are the exact solutions. Approximately the solutions are m = -.00328 and b= 99.5. Therefore your equation becomes T=-.00328A + 99.5. Double-checking by using your original values T(1381) = 95.0 and T(5951) = 80.0. These values match your data values so m and b are correct.
m=-.00328, b=99.5
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