f(x) = (x - 4)^2 - 5
Vertex (4 , -5)
This function opens upward and has min. value = -5
So range y >= - 5
So answer is A. -5 <= f(x) < ∞
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>
To solve, we use the volume of a square pyramid formula which is expressed as the product of the area of the base (a²) and the height divided by three(h/3).
Volume = a² (h/3)
1024 in³ = (a²) (12 in/3)
a² = 1024 in³ · 3 /12
a = √256 in²
a = 16 in
Thus, the length of the base is 16 in.
If he took 5 steps forward and then 5 steps backwards he would be at his starting position
