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vekshin1
3 years ago
10

Fb + 3b = d solve for b

Mathematics
1 answer:
jarptica [38.1K]3 years ago
4 0

Answer: the answer is d/f+3

sorry i couldnt find my work for this problem but i hope it helps !

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Expand to write an equivalent expression <br> -1/4 (-8x + 12y)
Tanzania [10]

The equivalent expression on expanding -1/4 (-8x + 12y) is 2x - 3y

<h3>Expansion of expression</h3>

Expansion of expression can be carried out using the distributive rule a shown:

A(B + C) = AB + AC

Note that <u>A is distributed over B and C</u>
Given the expression

-1/4 (-8x + 12y)

-1/4(-8x) - 1/4(12y)

2x - 3y

Hence the equivalent expression on expanding -1/4 (-8x + 12y) is 2x - 3y

Learn more on expansion here: brainly.com/question/26430239

#SPJ1

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Solucion de (2x)2-16x 32=98 con la formula general
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3 years ago
If |x|+10=1, then what does x equal
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Answer; there is literally no solution. it has to be a negative number.

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What is negative thirty divided by three
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The answer is -10

Step-by-step explanation:

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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}

6 0
2 years ago
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