3. To surf the internet for 15 minutes at an airport, it costs $\$ 4.05$. For 40 minutes, it costs $\$ 5.80$. Write and solve a linear equation to find the cost for surfing the web for one hour. 14. Water boils at $100^{\circ}$ Celsius or $212^{\circ}$ Fahrenheit. Water freezes at $0^{\circ}$ Celsius or $32^{\circ}$ Fahrenheit. If the weather forecaster says it will be $25^{\circ}$ Celsius today, write and solve a linear equation to find what Fahrenheit temperature this is.
Solution
Step 1 ：Let's denote the cost for surfing the internet for \(x\) minutes as \(C(x)\).
Step 2 ：We know that \(C(15) = \$4.05\) and \(C(40) = \$5.80\).
Step 3 ：We can write the equation of the line passing through the points \((15, 4.05)\) and \((40, 5.80)\).
Step 4 ：The slope of the line is \(\frac{5.80 - 4.05}{40 - 15} = \frac{1.75}{25} = \$0.07\) per minute.
Step 5 ：So, the equation of the line is \(C(x) = 0.07x + b\).
Step 6 ：To find \(b\), we can substitute one of the points into the equation. Let's use \((15, 4.05)\):
Step 7 ：\(4.05 = 0.07 \cdot 15 + b\),
Step 8 ：\(b = 4.05 - 0.07 \cdot 15 = \$2.95\).
Step 9 ：So, the cost for surfing the internet for \(x\) minutes is \(C(x) = 0.07x + 2.95\).
Step 10 ：To find the cost for surfing the web for one hour (or 60 minutes), we substitute \(x = 60\) into the equation:
Step 11 ：\(C(60) = 0.07 \cdot 60 + 2.95\)
Step 12 ：\(\boxed{C(60) = \$7.15}\)
Step 13 ：We know that \(f=32+1.8c\).
Step 14 ：To find the Fahrenheit temperature when it is \(25^\circ\) Celsius, we substitute \(c = 25\) into the equation:
Step 15 ：\(f =32+1.8c\)
Step 16 ：\(f =32+1.8(25)\)
Step 17 ：\(\boxed{f = 77^\circ}\)
