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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Answer: 2.1925153x10 to the 3rd
Step-by-step explanation: hope it helps
To factor, you can first treat it like a single bracket and find the common factor. In this case, the common factor is 3x, so you get
3x(x² + 7x + 12)
Now you can factor the bracket normally, by finding factors of 12 that add up to make 7. The factors would be 3 and 4, so the bracket becomes (x + 3)(x + 4).
This leaves your final answer as
3x(x + 3)(x + 4)
I hope this helps!
Answer:
wht cube
Step-by-step explanation: