namely, let's rationalize the denominator in the fraction, for which case we'll be using the <u>conjugate</u> of that denominator, so we'll multiply top and bottom by its <u>conjugate</u>.
so the denominator is 5 + i, simply enough, its conjugate is just 5 - i, recall that same/same = 1, thus (5-i)/(5-i) = 1, and any expression multiplied by 1 is just itself, so we're not really changing the fraction per se.
![\bf \cfrac{2}{5+i}\cdot \cfrac{5-i}{5-i}\implies \cfrac{2(5-i)}{\stackrel{\textit{difference of squares}}{(5+i)(5-i)}}\implies \cfrac{2(5-i)}{\stackrel{\textit{recall }i^2=-1}{5^2-i^2}}\implies \cfrac{2(5-i)}{25-(-1)} \\\\\\ \cfrac{2(5-i)}{25+1}\implies \cfrac{2(5-i)}{26}\implies \cfrac{5-i}{13}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B2%7D%7B5%2Bi%7D%5Ccdot%20%5Ccfrac%7B5-i%7D%7B5-i%7D%5Cimplies%20%5Ccfrac%7B2%285-i%29%7D%7B%5Cstackrel%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%285%2Bi%29%285-i%29%7D%7D%5Cimplies%20%5Ccfrac%7B2%285-i%29%7D%7B%5Cstackrel%7B%5Ctextit%7Brecall%20%7Di%5E2%3D-1%7D%7B5%5E2-i%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B2%285-i%29%7D%7B25-%28-1%29%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B2%285-i%29%7D%7B25%2B1%7D%5Cimplies%20%5Ccfrac%7B2%285-i%29%7D%7B26%7D%5Cimplies%20%5Ccfrac%7B5-i%7D%7B13%7D)
Answer: ![-4](https://tex.z-dn.net/?f=-4%3C%3D%20x%3C%3D%206)
Step-by-step explanation:
Domain: x-axis
Range: y-axis
The relation graph shows the points in which the domain is located. To find the Domain, go to the point in the x-axis (the horizontal line) and note the number where the point lies. For this question, the point on the left side of the graph lies on negative four (
), and on the right side, the point is on 6. Therefore, the domain of this relation is negative four is greater than/equal to x is less than/equal to six. It could also be written like this:
.
Learn more: brainly.com/question/24574301
Answer:
y = 2x+2
Step-by-step explanation:
y = mx+b
m = slope = 2
b = y-intercept = 2
Use calcutlator and it all smushed up i can barley see it