Question

1. let $f(x)=3 - x$ and $g(x)=7x + 2$. determine $(f - g)$.
2. which expression is equivalent to $\frac{sqrt{x}}{2sqrt{x}-3}$?

a $\frac{5sqrt{x}}{2x - 9}$
b $\frac{5sqrt{x}}{4x - 9}$
c $\frac{2x+3sqrt{x}}{2x - 9}$
d $\frac{2x+3sqrt{x}}{4x - 9}$

1. a cylindrical pipe is 9 ft long and has a volume of $100\\ \text{ft}^3$. find its approximate diameter to the nearest hundredth of a foot. hint: $v = \pi r^2h$

\\_\\_\\_\\_\\_\\_ ft

1. let $f(x)=1+\sqrt{x}$. find $f^{-1}(2)$.
2. solve $\sqrt{4x - 16}=\sqrt{x - 4}$.
3. the volume of a sphere is $v(r)=\frac{4}{3}\pi r^3$ and the radius is increasing 2 mm per second. the function $r(t)=2t$ gives the radius at time $t$ seconds. what is $v(r(t))$?
4. solve the equation $x + 3=\sqrt{x + 33}$.

Explanation:

Problem 10

Step1: Define $(f-g)(x)$

$(f-g)(x) = f(x) - g(x)$

Step2: Substitute given functions

$(f-g)(x) = (3 - x) - (7x + 2)$

Step3: Simplify the expression

$(f-g)(x) = 3 - x -7x -2 = 1 - 8x$

Step1: Rationalize the denominator

Multiply numerator/denominator by $2\sqrt{x}+3$:
$\frac{\sqrt{x}}{2\sqrt{x}-3} \cdot \frac{2\sqrt{x}+3}{2\sqrt{x}+3}$

Step2: Expand numerator and denominator

Numerator: $\sqrt{x}(2\sqrt{x}+3) = 2x + 3\sqrt{x}$
Denominator: $(2\sqrt{x})^2 - 3^2 = 4x - 9$

Step1: Rearrange volume formula for $r$

$V = \pi r^2 h \implies r^2 = \frac{V}{\pi h}$

Step2: Substitute $V=100, h=9$

$r^2 = \frac{100}{\pi \cdot 9} = \frac{100}{9\pi}$

Step3: Solve for radius $r$

$r = \sqrt{\frac{100}{9\pi}} = \frac{10}{3\sqrt{\pi}} \approx \frac{10}{3 \cdot 1.772} \approx 1.88$

Step4: Calculate diameter ($d=2r$)

$d = 2 \cdot 1.88 = 3.76$

Answer:

$\boldsymbol{1 - 8x}$

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Problem 11