Easy 7th grade work sheet ! Please help , I’ll give brainliest (all answers)

Mathematics

1 answer:

jeka57 [31]3 years ago

4 0

1. +16
2. -11
3. -6
4. 9
5. 8
6. 5
7. Simply place dots on the numbers stated, which are 11, -5, and -8.
8. The absolute value of a non-zero number is positive because absolute value is a magnitude, and is either positive or zero. The absolute value of any non-zero number can be thought of as its distance from zero, which will always be positive.

Please mark as branliest if i helped!!:)

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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 )

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