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2 years ago

Please help Solve for x and y 60 points!!!!

Mathematics

2 answers:

dlinn [17]2 years ago

6 0

Answer:

see explanation

Step-by-step explanation:

(1)

x + 20 and 45 are vertically opposite angles and are congruent, so

x + 20 = 45 ( subtract 20 from both sides )

x = 25

Similarly 5y and 135 are vertically opposite angles and are congruent

5y = 135 ( divide both sides by 5 )

y = 27

(2)

x + 30 and 70 are vertically opposite angles and are congruent, then

x + 30 = 70 ( subtract 30 from both sides )

x = 40

Similarly y + 50 and 110 are vertically opposite and congruent , so

x + 50 = 110 ( subtract 50 from both sides )

x = 60

Neko [114]2 years ago

3 0

Answer:

60 point 60 point 60 point

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In Salem, 4 of the 10 radio stations play classical music.

aliina [53]

Answer:

40% or 4/10 (assuming you just turned on the radio)

Step-by-step explanation:

If you just turned on the radio, it would be 4/10. If you changed from a channel that wasn't playing classical music then it would be 4/9. If you changed from a channel that was playing classical music then it would be 3/9.

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2 years ago

In 2015, there were roughly 1x10^6 high school football players and 2x10^3 professional football players in the United States.

melisa1 [442]

Answer:

500.

Step-by-step explanation:

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3 years ago

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 )

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