To complete each ordered pair, you would have to substitute the 1 and 2 in for y. Let's start with the 2 first. X-2(2)=7, x-4=7. Add 4 on each side. X=11. First ordered pair is (11,2). Now do the 1. X-2(1)=7, x-2=7. Add 2 on both sides. Second ordered pair is (9,1). Answers are (11,2) and (9,1)
When evaluating b^2c^-1 for b=8 and c=4, the answer is -16
E = b^2c–1
x-a = 1/xa
E = b^2c - 1
= 8^2 x (-4)-1
= 64/-4
= -16
Answer:
3, 6, 7, 9
Step-by-step explanation:
We can take a look at each value from 1 to 10
We see that for 3, 6, 7, and 9 the decimal repeats forever for some numerators. For example, 1/3 repeats forever, 4/6 repeats forever, 5/7 repeats forever, 1/9 repeats forever.
<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>
