Question

a baker has a bin filled with 30 cups of flour. his signature cake requires one and a half cups of flour. determine which graph and which equation represent the amount of flour in the bin, f, after he bakes c signature cakes. graphs (with four graphs showing f vs c, two decreasing from (0,30) and two increasing, with axes labeled f (30,24,18,12,6) and c (0,4,8,12,16,20))

Explanation:

Step1: Define the relationship

The initial amount of flour is 30 cups. Each cake uses \(1\frac{1}{2}=\frac{3}{2}\) cups of flour. So the amount of flour remaining \(F\) after baking \(c\) cakes is a linear relationship: \(F = 30-\frac{3}{2}c\).

Step2: Analyze the graph characteristics

• The equation is in the form \(y = mx + b\) (here \(F = -\frac{3}{2}c+30\)), so the slope \(m = -\frac{3}{2}\) (negative, so the line should be decreasing) and the y - intercept (when \(c = 0\)) is 30.
• For the x - intercept (when \(F = 0\)):

Set \(0=30-\frac{3}{2}c\)
\(\frac{3}{2}c=30\)
\(c = 30\times\frac{2}{3}=20\)? Wait, no, wait: Wait, \(\frac{3}{2}c = 30\) gives \(c=\frac{30\times2}{3}=20\)? Wait, but let's check the first graph: when \(c = 0\), \(F = 30\) (correct y - intercept), when \(F = 0\), \(c=\frac{30}{\frac{3}{2}}=20\)? Wait, the first graph has \(c\) - intercept at 16? Wait, maybe I miscalculated. Wait, \(1\frac{1}{2}=\frac{3}{2}\), so the number of cakes \(c\) that can be made is \(\frac{30}{\frac{3}{2}}=20\). Wait, maybe the first graph (top - left) has a mistake? Wait, no, let's recalculate: if each cake uses \(1.5\) cups, then for \(c = 0\), \(F = 30\). For \(c = 4\), \(F=30 - 1.5\times4=30 - 6 = 24\) (matches the top - left graph: when \(c = 4\), \(F = 24\)). For \(c = 8\), \(F=30-1.5\times8 = 30 - 12 = 18\) (matches). For \(c = 12\), \(F=30 - 1.5\times12=30 - 18 = 12\) (matches). For \(c = 16\), \(F=30-1.5\times16 = 30 - 24 = 6\) (matches). For \(c = 20\), \(F=30-1.5\times20=30 - 30 = 0\). Wait, the first graph (top - left) has \(F = 0\) at \(c = 16\), which is wrong. Wait, maybe the problem's graph is mis - drawn, but according to the equation \(F = 30-1.5c\), the correct graph should have a y - intercept of 30, a negative slope, and x - intercept at \(c = 20\). But among the given graphs, the top - left graph has \(F(0)=30\), decreasing, and let's check the values: at \(c = 4\), \(F = 24\) (30 - 6), at \(c = 8\), \(F = 18\) (30 - 12), at \(c = 12\), \(F = 12\) (30 - 18), at \(c = 16\), \(F = 6\) (30 - 24), at \(c = 20\), \(F = 0\) (30 - 30). Wait, maybe the x - axis in the first graph is labeled up to 20, but the line ends at \(c = 16\) with \(F = 0\)? No, maybe I made a mistake. Wait, \(1.5\times16 = 24\), \(30 - 24 = 6\), no, \(1.5\times16=24\), \(30 - 24 = 6\), so when \(c = 16\), \(F = 6\), not 0. Wait, I think I messed up the x - intercept calculation. Let's do it again: \(F=30 - 1.5c\). Set \(F = 0\): \(1.5c=30\) \(\Rightarrow\) \(c = 20\). So the correct graph should have \(F = 0\) at \(c = 20\). But among the given graphs, the top - left graph has a decreasing line with \(F(0)=30\), and let's check the slope: from \(c = 0,F = 30\) to \(c = 4,F = 24\), the change in \(F\) is \(24 - 30=-6\), change in \(c\) is \(4 - 0 = 4\), slope is \(\frac{-6}{4}=-\frac{3}{2}\) (correct). So the equation is \(F = 30-1.5c\) (or \(F = 30-\frac{3}{2}c\)) and the graph should be the one with a decreasing line starting at \((0,30)\) and having a slope of \(-\frac{3}{2}\).

Answer:

The equation is \(F = 30-\frac{3}{2}c\) (or \(F = 30 - 1.5c\)) and the graph should be the one with a decreasing line (negative slope) with \(F(0)=30\) (the top - left graph among the given graphs, assuming the x - intercept calculation was mis - visualized in the graph, but based on the slope and y - intercept, the equation is \(F=30 - 1.5c\) and the graph is the decreasing line starting at (0,30)).