[tex}r - 5 \frac{5}6 + 5\frac{5}6 = 10 +5\frac{5}6{/tex]
<em>Given - a+b+c = 0</em>
<em>To prove that- </em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
<em>Now we know that</em>
<em>when x+y+z = 0,</em>
<em>then x³+y³+z³ = 3xyz</em>
<em>that means</em>
<em> (x³+y³+z³)/xyz = 3 ---- eq 1)</em>
<em>Lets solve for LHS</em>
<em>LHS = a²/bc + b²/ac + c²/ab</em>
<em>we can write it as LHS = a³/abc + b³/abc + c</em><em>³</em><em>/abc</em>
<em>by multiplying missing denominators,</em>
<em>now take common abc from denominator and you'll get,</em>
<em>LHS = (a³+b³+c³)/abc --- eq (2)</em>
<em>Comparing one and two we can say that</em>
<em>(a³+b³+c³)/abc = 3</em>
<em>Hence proved,</em>
<em>a²/bc + b²/ac + c²/ab = 3</em>
1 Simplify \frac{4}{15}x
15
4
x to \frac{4x}{15}
15
4x \frac{4x}{15}=1.44
4x =1.44
2 Multiply both sides by 1515.
4x=1.44\times 15
4x=1.44×15
3 Simplify 1.44\times 151.44×15 to 21.621.6.
4x=21.6
4x=21.6
4 Divide both sides by 44.
x=\frac{21.6}{4}
x= 4
21.65 Simplify \frac{21.6}{4} 421.6 to 5.45.4.
x=5.4
x=5.4
Answer: 148
Work: 50 men multiplied by 3 =150
If there is two less women from 3 multiplied by 50 (150) , then 150 - 2 = 148
