Answer:
an open circle on -1.3 on the line and draw a line to the right with an arrow.
Step-by-step explanation:
On your number line, draw an open circle on -1.3 and then draw a line on the number line going to the right and ending with an arrow at the end. The open circle means that it is that starting point, but is not equal to it.
We will use demonstration of recurrences<span>1) for n=1, 10= 5*1(1+1)=5*2=10, it is just
2) assume that the equation </span>10 + 30 + 60 + ... + 10n = 5n(n + 1) is true, <span>for all positive integers n>=1
</span>3) let's show that the equation<span> is also true for n+1, n>=1
</span><span>10 + 30 + 60 + ... + 10(n+1) = 5(n+1)(n + 2)
</span>let be N=n+1, N is integer because of n+1, so we have
<span>10 + 30 + 60 + ... + 10N = 5N(N+1), it is true according 2)
</span>so the equation<span> is also true for n+1,
</span>finally, 10 + 30 + 60 + ... + 10n = 5n(n + 1) is always true for all positive integers n.
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The first option is how you wrote your statement in math.
So you for the answer you have to add all alike-variables, which all of them are alike in this question. 5c+8c=13c. 13c+3c= 16c, so 16c would be your answer
If you take the square root of integers 1-100, the only rational numbers will be the perfect squares, hence the square root. Count how many squares there are between 1 and 100, inclusive, and subtract that number from 100 to get the number of rational and irrational numbers.
