keeping in mind that perpendicular lines have negative reciprocal slopes, hmmmm what's the slope of the equation above anyway?
![\bf x+y=6\implies y = \stackrel{\stackrel{m}{\downarrow }}{-1}x+6\qquad \impliedby \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} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20x%2By%3D6%5Cimplies%20y%20%3D%20%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-1%7Dx%2B6%5Cqquad%20%5Cimpliedby%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

so we're really looking for the equation of a line whose slope is 1 and runs through (-5,-6).
![\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-6})~\hspace{10em} \stackrel{slope}{m}\implies 1 \\\\\\ \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}{(-6)}=\stackrel{m}{1}[x-\stackrel{x_1}{(-5)}] \\\\\\ y+6=1(x+5)\implies y+6=x+5\implies y=x-1](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B-5%7D~%2C~%5Cstackrel%7By_1%7D%7B-6%7D%29~%5Chspace%7B10em%7D%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%201%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B%28-6%29%7D%3D%5Cstackrel%7Bm%7D%7B1%7D%5Bx-%5Cstackrel%7Bx_1%7D%7B%28-5%29%7D%5D%20%5C%5C%5C%5C%5C%5C%20y%2B6%3D1%28x%2B5%29%5Cimplies%20y%2B6%3Dx%2B5%5Cimplies%20y%3Dx-1)
You're given an angle, the side adjacent to the angle, and are looking for the hypotenuse.
Use cosine = adjacent / hypotenuse.
cos(60) = 2√(3) / x
Rearrange the equation to solve for x (multiply both sides by x, then divide both sides by cos(60)).
x = (2√(3)) / (cos(60))
x equates to about 6.928
6 x 4 is the same as 6 divided by 1/4
the answer is :
1.5
Logarithmic functions and exponential functions are inverse and opposite of one another
The logarithmic function is
, and the length at 8 pascals is 3 units
<h3>How to determine the
logarithmic function</h3>
The exponential function is given as:

Express f(x) as y

Swap the positions of x and y

Take the logarithm of both sides

Apply the rule of logarithm

Divide both sides by log(2)

Apply the change of base rule of logarithm

When the strength is 8 pascals, we have:

Express 8 as 2^3

So, we have:

Evaluate log 2 base 2

Hence, the logarithmic function is
, and the length at 8 pascals is 3 units
Read more about logarithmic and exponential functions at:
brainly.com/question/11464095