5(b - 0.18b).....use the distributive property
5b - 0.90b <== ur equivalent expression
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)
V=1/3Bh B=area of base h=height
Answer:
15 weeks
Step-by-step explanation:
850-250= 600
40*15= 600
<span>We want to check how many intersections line A and B have, that is, we want to check how many common solutions do these equations have:
</span>
i) 2x + 2y = 8
ii) x + y = 4
<span>
use equation ii) to write y in terms of x as : y=4-x,
substitute y =4-x in equation i):
</span>2x + 2y = 8
2x + 2(4-x) = 8
<span>2x+8-2x=8
8=8
this is always true, which means the equations have infinitely many common solutions.
Answer: </span><span>There are infinitely many solutions.</span><span>
</span>
