The distance covered by the hare and the tortoise in t seconds are 8t and 5t respectively. (Simple Speed-Distance-Time relation)
The tortoise gets a 510m headstart so at t=0 is 1490m.
The functions representing the distance of both of them from the finish line is,
F(x)=2000-8t,. for hare
G(x)=1490-5t,. for tortoise
Answer:
49%
Step-by-step explanation:
to get the percentage you put the number of occupied seats over total number of seats and multiply it by 100 over 1
65 because if you put all the numbers in order and work out which one is in the middle you get the median. i enjoy using the rhyme. Hey diddle diddle, the medians the middle, you add then divide for the mean, the mode is the one that you see the most and the range is the difference between.
Answer:
x/5 - 7 = 8
Step-by-step explanation:
<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>
