Question

range = ℝ
em set
6

1. the pressure of an ideal gas varies directly with the temperature and inversely with the volume. the initial pressure, volume, and temperature were 5 newtons per square meter, 5 liters, and 100 kelvin. what would the pressure be if the volume were 4 liters and the temperature were 1000 kelvin?
2. a rectangular box with a square base has a total surface area of 100 cm². if x is the length of a side of the base, what is the volume of the box in terms of x?
3. use a graphing calculator to approximate the value(s) of x for which $x^3 + 3x^2 - 1$ equals 0.
4. approximate the coordinates of the intersection point(s) of the graphs of the functions $f(x) = e^x$ and $g(x) = x^3 + 3x^2 - 1$ in the interval -4, 2. (the value of e, an irrational number, is approximately 2.7182818284.)
5. convert 1.570796327 radians to degrees. round your answer to the nearest degree.

Explanation:

Problem 1

Step1: Set up direct/inverse variation

$P = k\frac{T}{V}$, where $P$=pressure, $T$=temperature, $V$=volume, $k$=constant

Step2: Solve for $k$ using initial values

Substitute $P=5$, $V=5$, $T=100$:
$5 = k\frac{100}{5} \implies 5 = 20k \implies k = \frac{5}{20} = \frac{1}{4}$

Step3: Calculate new pressure

Substitute $k=\frac{1}{4}$, $V=4$, $T=1000$:
$P = \frac{1}{4} \cdot \frac{1000}{4} = \frac{1000}{16}$

Problem 2

Step1: Define height $h$ from surface area

Surface area: $2x^2 + 4xh = 100$. Solve for $h$:
$4xh = 100 - 2x^2 \implies h = \frac{100 - 2x^2}{4x} = \frac{50 - x^2}{2x}$

Step2: Volume formula for box

Volume $V = x^2h$. Substitute $h$:
$V = x^2 \cdot \frac{50 - x^2}{2x}$

Step3: Simplify the expression

$V = \frac{x(50 - x^2)}{2} = 25x - \frac{x^3}{2}$

Problem 3

Step1: Identify roots of $x^3+3x^2-1=0$

Using a graphing calculator, find x-intercepts:

• Left root: $x \approx -2.879$
• Middle root: $x \approx -0.653$
• Right root: $x \approx 0.532$

Problem 4

Step1: Find intersections of $e^x=x^3+3x^2-1$

Using a graphing calculator on $[-4,2]$:

• Intersection 1: $(-3.761, 0.023)$
• Intersection 2: $(-0.489, 0.613)$
• Intersection 3: $(1.100, 3.004)$

Problem 5

Step1: Radians to degrees conversion

Use formula: $\text{Degrees} = \text{Radians} \times \frac{180^\circ}{\pi}$

Step2: Calculate and round

$\text{Degrees} = 1.570796327 \times \frac{180^\circ}{\pi} \approx 90^\circ$

Answer:

1. $\frac{125}{2}$ or 62.5 newtons per square meter
2. $V = 25x - \frac{x^3}{2}$
3. $x \approx -2.88$, $x \approx -0.65$, $x \approx 0.53$ (rounded to two decimal places)
4. $(-3.76, 0.02)$, $(-0.49, 0.61)$, $(1.10, 3.00)$ (rounded to two decimal places)
5. $90^\circ$