Jonas has 1728 copies of a 1 × 1 × 1 cube with the net shown, where c is a positive integer and c < 100. Using
these 1728 cubes, Jonas builds a large 12×12×12 cube in
such a way that the sum of the numbers on the exterior
faces is as large as possible. For some values of c, the sum
of the numbers on the exterior faces is between 80 000 and
85 000. The number of such values of c is
(A) 39 (B) 38 (C) 37
(D) 36 (E) 35

10x

step-by-step explanation:

the greatest common factor 30 and 20 is 10

5 is a common factor of 30 and 20 but 10 is also a factor and 10 is greater than 5 so 5 will definitely not be apart of our answer

now the most amount of x's you factor our is just 1 factor of 1

so 10x the gcf of the terms in the polynomial

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.