Can someone help me with these questions please, ill give you brainliest.

Mathematics

1 answer:

lesya692 [45]3 years ago

5 0

Problem 9.

Using the Angle Addition Postulate,

m<QRS + m<SRT = m<QRT

Now we replace the measures of the three angles above with the value or expressions we are given, and we solve for x.

2x + 8 + 3x + 14 = 82

5x + 22 = 82

5x = 60

x = 12

m<QRS = 2x + 8 = 2(12) + 8 = 24 + 8 = 32

Answer: 32 degrees

Problem 10.

Using the Angle Addition Postulate,

m<ABC + m<CBD = m<ABD

Now we replace the measures of the two angles above with the values we are given, and solve for the unknown angle.

115 + m<CBD = 180

m<CBD = 180 - 115

m<CBD = 65

Answer: m<CBD = 65 degrees

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Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).

Svetradugi [14.3K]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5$

well then therefore

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}$

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}$