now, let's take a peek at the denominators, we have 3 and 8 and 12, we can get an LCD of 24 from that.
Let's multiply both sides by the LCD of 24, to do away with the denominators.
so, let's recall that a whole is "1", namely 500/500 = 1 = whole, or 5/5 = 1 = whole or 24/24 = 1 = whole. So the whole class will yield a fraction of 1/1 or just 1.
![\bf ~\hspace{7em}\stackrel{\textit{basketball}}{\cfrac{1}{3}}+\stackrel{\textit{soccer}}{\cfrac{1}{8}}+\stackrel{\textit{football}}{\cfrac{5}{12}}+\stackrel{\textit{baseball}}{x}~=~\stackrel{\textit{whole}}{1} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{24}}{24\left(\cfrac{1}{3}+\cfrac{1}{8}+\cfrac{5}{12}+x \right)=24(1)}\implies (8)1+(3)1+(2)5+(24)x=24 \\\\\\ 8+3+10+24x=24\implies 21+24x=24\implies 24x=3 \\\\\\ x=\cfrac{3}{24}\implies x=\cfrac{1}{8}](https://tex.z-dn.net/?f=%5Cbf%20~%5Chspace%7B7em%7D%5Cstackrel%7B%5Ctextit%7Bbasketball%7D%7D%7B%5Ccfrac%7B1%7D%7B3%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bsoccer%7D%7D%7B%5Ccfrac%7B1%7D%7B8%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bfootball%7D%7D%7B%5Ccfrac%7B5%7D%7B12%7D%7D%2B%5Cstackrel%7B%5Ctextit%7Bbaseball%7D%7D%7Bx%7D~%3D~%5Cstackrel%7B%5Ctextit%7Bwhole%7D%7D%7B1%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B24%7D%7D%7B24%5Cleft%28%5Ccfrac%7B1%7D%7B3%7D%2B%5Ccfrac%7B1%7D%7B8%7D%2B%5Ccfrac%7B5%7D%7B12%7D%2Bx%20%5Cright%29%3D24%281%29%7D%5Cimplies%20%288%291%2B%283%291%2B%282%295%2B%2824%29x%3D24%20%5C%5C%5C%5C%5C%5C%208%2B3%2B10%2B24x%3D24%5Cimplies%2021%2B24x%3D24%5Cimplies%2024x%3D3%20%5C%5C%5C%5C%5C%5C%20x%3D%5Ccfrac%7B3%7D%7B24%7D%5Cimplies%20x%3D%5Ccfrac%7B1%7D%7B8%7D)
X=-6
You plus in x as -6 and you already know that y=5x and you plug in x=-6
It should look like 2(-6)-5(-6)=18
Yes they are if that helps, if you still are unsure than just mesure them and you will see
80% of 1000 is 800 hope this helps :)
I believe the correct answer is false. Quadratic equations that can be solved using the quadratic formula cannot always be solved by factoring. <span>Not </span>all quadratic equations can<span> be </span>solved by factoring<span>. Another method that can be used is completing the square. Hope this answers the question.</span>
