The equation is w = 3d. For a distance of d = 2 units, substituting it into the linear equation gives the work done as w = 3(2) = 6 units. Plotting (2, 6) confirms this on the graph.
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Let the tens digit be x and units digit be x – 3. The number is 11x – 3. Interchanging digits gives 11x – 30. Adding them, 22x – 33 = 143, solving to x = 8. The numbers are ...
Let the unknown number be x. The problem gives the linear equation 5/2 x + 2/3 = -7/12. Shifting 2/3 over gives 5/2 x = -15/12, which simplifies to the final number x = -1/2.
Using the relation C = aF + b, substitute the given conditions. From the melting point, 0 = 32a + b. From the boiling point, 100 = 212a + b. Solving this gives a = 5/9 and b = -160/9.
Substitute the values into y = ax + b. For 10 modules, 400 = 10a + b. For 14 modules, 500 = 14a + b. Subtracting the equations gives 4a = 100, which results in a = 25 and b ...