Write a recrusive rule for the exponential functionf(x)=0.25(2/3)^x
1 answer:
Answer:
f(0) = 0.25 = 1/4
f(x) = f(x-1) × 2/3
Step-by-step explanation:
a0 = 0.25×(2/3)⁰ = 0.25×1 = 0.25 = 1/4
a1 = 0.25×(2/3)¹ = 0.25×(2/3) = 1/4 × 2/3 = 2/12 = 1/6 =
= a0 × 2/3
a2 = 0.25×(2/3)² = 0.25×(4/9) = 1/4 × 4/9 = 4/36 = 1/9 =
= a1 × 2/3
=>
an = an-1 × 2/3, n >= 1
but what about negative x ?
is a0 = a-1 × 2/3 ?
so, yes, a0 = a-1 × (2/3)
and a-1 = a-2 × (2/3)
...
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That's the cross product in the purest form. When we're away from the origin, a arbitrary triangle with vertices will have the same area as one whose vertex C is translated to the origin.
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That's a perfectly useful formula right there. But it's usually multiplied out:
That's the usual form, the sum of cross products. Let's line up our numbers to make it easier.
(1, 2), (3, 4), (−7, 7)
(−7, 7),(1, 2), (3, 4),
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