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Answer:
-10
Step-by-step explanation:
Let x = number of postcards sold (as this is the independent variable)
Let y = profit (as this is the dependent variable)
We need to find the equation of the straight line that passes through the two points.
Find the gradient using the formula: [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
where [tex]m[/tex] is the slope and [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] are the two points
Let [tex](x_1, y_1)[/tex] = (4, 2) and [tex](x_2, y_2)[/tex] = (10, 20)
Therefore, [tex]m=\frac{20-2}{10-4}=\frac{18}{6} =3[/tex]
Now use the equation of line formula: [tex]y-y_1=m(x-x_1)[/tex]
y - 2 = 3(x - 4)
y - 2 = 3x - 12
y = 3x - 10
If Max doesn't sell any postcards, then x = 0.
Substituting x = 0 into the equation of the line: y = (3 x 0) - 10 = -10
Therefore, Max's profit prior to selling any postcards is -10 (so he is at a net loss of 10 before selling any postcards).