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GarryVolchara [31]
3 years ago
9

Factor 28+56t+28w28+56t+28w28, plus, 56, t, plus, 28, w to identify the equivalent expressions. Choose 2 answers:

Mathematics
1 answer:
Effectus [21]3 years ago
6 0

Answer: The factorization of  28+56t+28w =28(1+2t+w)

The factors of 28+56t+28w are 28 and 1+2t+w.

Step-by-step explanation:

The given expression : 28+56t+28w

To factorize it, we need to find the common factor.

As 56 can be written as 2 x 28.

So, the above expression would become

28+2\times28t+28w

Now, taking 28 as common from all the terms, we will get

28(1+2t+w)

Thus, the factorization of  28+56t+28w =28(1+2t+w)

And the factors of  28+56t+28w are 28 and 1+2t+w.

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

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Since these are solutions

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