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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</span>
Answer:
B
Step-by-step explanation:
We can notice that based on the given pattern, the average rate of change from to , where , is .
So, the average rate of change from to is .
the answer would be revenue
Answer:
In a quadratic equation of the shape:
y = a*x^2 + b*x + c
we hate that the discriminant is equal to:
D = b^2 - 4*a*c
This thing appears in the Bhaskara's formula for the roots of the quadratic equation:
You can see that the determinant is inside a square root, this means that if D is smaller than zero we will have imaginary roots (the graph never touches the x-axis)
If D = 0, the square root term dissapear, and this implies that both roots of the equation are the same, this means that the graph touches the x axis in only one point, wich coincides with the minimum/maximum of the graph)
If D > 0 we have two different roots, so the graph touches the x-axis in two different points.