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3 years ago

Please Help! :) Solve for X. The polygons in each pair are similar.

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

8 0

Answer:

x=2.65

Step-by-step explanation:

First, we need to figure out how much the length of the sides is changing between the equations. In this case it is changing by a factor of 2.5. Now we need to work backwards. First use this equation:

$4x-5*2.5=14$

This is because whatever the length of the side with x is, will be 14/2.5. So divide both sides by 2.5:

$4x-5=5.6$

Next, add 5 to both sides:

$4x=10.6$

Then divide both sides by 4:

$x=2.65$

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The sum of a number times 3 and 7 equals 2

ella [17]

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3 x = 2 - 7
3 x = - 5
x = - 5/3

3 ( -5/3) + 7 = - 15/3 + 7 = - 5 + 7 = 2

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3 years ago

How do you find a vector that is orthogonal to 5i + 12j ?

Rashid [163]

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\$

$\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
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\\\\\\
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if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
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\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

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