You have to multiply and add!
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 GCF of 33c and 55cd is
33c/11c = 3
55cd/11c = 5d
the greatest common factor is 11c
hope this helps
Answer:
1, 5 and 6.
Step-by-step explanation:
The sides of this triangles has lengths in the ratio x : √3x : 2x where 2x is the hypotenuse.
When C = 30, the side AB is the small leg (=x).
When B = 90, the hypotenuse is AC ( =2x).
When A = 60, the larger leg is BC (=√3x).
1. AB = 4, BC = 4√3 :- Are possible legs.
5 AB = 7, AC = 14. :- AB = small leg, AC = hypotenuse.
6. AB = 11, BC = 11√3:- AB = small, BC = larger leg.