$\bf \stackrel{x-intercept}{(\stackrel{x_1}{2}~,~\stackrel{y_1}{0})}\qquad \stackrel{y-intercept}{(\stackrel{x_2}{0}~,~\stackrel{y_2}{6})} \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-0}{0-2}\implies \cfrac{6}{-2}\implies -3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=-3(x-2)\implies y=-3x+6$

5 0

2 years ago

Use the figure to the right to find the missing lengths of the triangle.

pychu [463]

To find AB(x) and BC(y), you can do(there are multiple ways you can do this):

tan A = $\frac{y}{7}$

tan 60° = $\frac{y}{7}$ or (tan 60°) · 7 = y

tan 60° = $\frac{\frac{\sqrt{3} }{2} }{ \frac{1}{2} } = \sqrt{3}$

√3 · 7 = y

7√3 cm = y

sin B = $\frac{7}{x}$

sin 30° = $\frac{7}{x}$ or $x= \frac{7}{sin 30}$

sin 30° = $\frac{\frac{1}{2} }{1}$ = $\frac{1}{2}$

x = $\frac{7}{\frac{1}{2} } = 14$

x = 14 cm

AB = 14cm

BC = 7√3 cm

8 0

3 years ago

Find cos(2theta) if cos(theta)= 3/4

yuradex [85]

Sin theta = 8/17 and theta is in the second quadrant. Find sin(2theta),cos(2theta),tan(2theta) exactly sin(2theta) cos(2theta) tan(2theta) sin(2theta) would it be 2 x (8/17) cos(2theta) would be 2 x (15/17) tan(2theta) would be 2 x (8/17 divided by 15/17) is this correct?

6 0

3 years ago