Determine two consecutive odd integers such that the sum of their squares is a minimum.
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Answer:
Step-by-step explanation:
We have a circle that is split in three sections, two of which we know and we are asked to find the third missing section.
For the circle, we know that 4/5 and 1/10 is fit. Now we need the last one, to solve, we need to get the same denominator and see how much is missing.
Since 1/10 is our highest denominator, let's change 4/5 to have 10 as a denominator. Which would be through multiplying 5 to get 10.
What times 5 equals 10?
2
Now multiply both numerator and denominator by 2 to get our portion.
Now we have the same denominator, let's add our two fractions and see how much we have left.
8/10 + 1/10
9/10
We have 1/10 missing, therefore 1/10 is the answer.
APEX ! Non-linear systems of equation.
