3(x-8)=12
I got this equation since there are three of the "x-8" sections that add up to the twelve length bar above
(24, 12) and (36, 0). The least amount of flowering plants occurs when x=2y, and the largest amount occurs when y=0. These two points satisfy both conditions and both sum to 36.
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.
<span>
</span>
Answer:
Perimeter of the polygon p = 24
Step-by-step explanation:
Step(i):-
Given the vertices of a polygon are
P(0,4) ,Q( 5,4) ,R( 5,-3) and S(0,-3)
The distance of PQ
The distance of QR
The distance of RS
The distance of PS
<u><em>Step(ii):-</em></u>
Perimeter of the polygon
= sum of all sides of polygon
p = a+ b+ c+ d
p = 5+7+5+7
p = 24
<u><em>Final answer:-</em></u>
Perimeter of the polygon p = 24
