13. use the law of sines to find how many triangles, abc, have the following parameters: α = 30°, c = 49.2…

13.

Explanation:

Step1: Apply Law of Sines

By the Law of Sines, $\frac{\sin\alpha}{a}=\frac{\sin\gamma}{c}$. Substitute $\alpha = 30^{\circ},c = 49.2,a = 24.6$. So $\sin\gamma=\frac{c\sin\alpha}{a}=\frac{49.2\times\sin30^{\circ}}{24.6}$.

Step2: Calculate $\sin\gamma$

$\sin\gamma=\frac{49.2\times\frac{1}{2}}{24.6}=\frac{24.6}{24.6}=1$. Since $\sin\gamma = 1$, $\gamma = 90^{\circ}$, and there is one - triangle.

Answer:

One triangle

14.

Explanation:

Step1: Rewrite $\frac{5\pi}{12}$

$\frac{5\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6}$. Then $\sin\frac{5\pi}{12}=\sin(\frac{\pi}{4}+\frac{\pi}{6})$.

Step2: Apply sum - of - angles formula

$\sin(A + B)=\sin A\cos B+\cos A\sin B$. Here $A=\frac{\pi}{4},B = \frac{\pi}{6}$. So $\sin(\frac{\pi}{4}+\frac{\pi}{6})=\sin\frac{\pi}{4}\cos\frac{\pi}{6}+\cos\frac{\pi}{4}\sin\frac{\pi}{6}$.

Step3: Substitute values

$\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2},\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2},\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2},\sin\frac{\pi}{6}=\frac{1}{2}$. Then $\sin(\frac{\pi}{4}+\frac{\pi}{6})=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$.

Answer:

$\frac{\sqrt{6}+\sqrt{2}}{4}$

15.

Explanation:

Step1: Rearrange the equation

Given $\sin^{2}(x)-\frac{1}{2}=0$, then $\sin^{2}(x)=\frac{1}{2}$, so $\sin(x)=\pm\frac{\sqrt{2}}{2}$.

Step2: Find $x$ in the given interval

For $\sin(x)=\frac{\sqrt{2}}{2}$ and $x\in(0,\pi)$, $x = \frac{\pi}{4}$ or $x=\frac{3\pi}{4}$.

Answer:

$\left{\frac{\pi}{4},\frac{3\pi}{4}\right}$

16.

Explanation:

Step1: Find the amplitude

For the function $y = A\cos(Bx)$, the amplitude is $|A|$. For $f(x)=\frac{1}{2}\cos(3x)$, $A=\frac{1}{2}$, so the amplitude is $\frac{1}{2}$.

Step2: Find the period

The period of $y = A\cos(Bx)$ is $T=\frac{2\pi}{|B|}$. Here $B = 3$, so $T=\frac{2\pi}{3}$.

Step3: Find the domain

The domain of $y=\cos(3x)$ and thus of $y=\frac{1}{2}\cos(3x)$ is all real numbers, $\mathbb{R}$.

Step4: Find the range

Since $- 1\leqslant\cos(3x)\leqslant1$, then $-\frac{1}{2}\leqslant\frac{1}{2}\cos(3x)\leqslant\frac{1}{2}$, so the range is $\left[-\frac{1}{2},\frac{1}{2}\right]$.

Answer:

Amplitude: $\frac{1}{2}$, Period: $\frac{2\pi}{3}$, Domain: $\mathbb{R}$, Range: $\left[-\frac{1}{2},\frac{1}{2}\right]$

17.

Explanation:

Step1: Use logarithm properties

Since $\ln a+\ln b=\ln(ab)$ and $\ln1 = 0$, the equation $\ln4x=\ln1+\ln(3x + 1)$ becomes $\ln4x=\ln(3x + 1)$.

Step2: Remove the logarithms

If $\ln a=\ln b$, then $a = b$. So $4x=3x + 1$.

Step3: Solve for $x$

Subtract $3x$ from both sides: $4x-3x=3x + 1-3x$, getting $x = 1$.

Answer:

$x = 1$

18.

To graph $g(x)=\log_{2}(x)-3$:

• The domain of $y = \log_{2}(x)$ is $x>0$.
• The vertical asymptote is $x = 0$.
• When $x = 1$, $g(1)=\log_{2}(1)-3=0 - 3=-3$.
• The graph of $y=\log_{2}(x)$ is shifted down 3 units.

19.

To graph $f(x)=\frac{x^{2}-3x + 2}{x - 3}=\frac{(x - 1)(x - 2)}{x - 3}$:

• The domain is $x\neq3$.
• Vertical asymptote: $x = 3$.
• To find the $x$ - intercepts, set $y = 0$, then $(x - 1)(x - 2)=0$, so $x=1$ and $x = 2$ are the $x$ - intercepts.
• To find the $y$ - intercept, set $x = 0$, then $y=\frac{2}{-3}=-\frac{2}{3}$.
• We can also use long - division or rewrite it as $f(x)=x+\frac{2}{x - 3}$.

20.

Explanation:

Step1: Since $x=-4$ is a double root

$(x + 4)^{2}=x^{2}+8x + 16$ is a factor of $P(x)=x^{4}+8x^{3}+19x^{2}+24x + 48$.

Step2: Perform polynomial long - division

Divide $x^{4}+8x^{3}+19x^{2}+24x + 48$ by $x^{2}+8x + 16$. We get $x^{4}+8x^{3}+19x^{2}+24x + 48=(x^{2}+8x + 16)(x^{2}+3)$.

Step3: Find the roots

For $x^{2}+8x + 16=(x + 4)^{2}=0$, $x=-4$ with multiplicity 2. For $x^{2}+3=0$, $x=\pm\sqrt{-3}=\pm i\sqrt{3}$ with multiplicity 1.

Answer:

$x=-4$ (multiplicity 2), $x = i\sqrt{3}$ (multiplicity 1), $x=-i\sqrt{3}$ (multiplicity 1)