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Rashid [163]
3 years ago
9

#7 What is f(4) if f(1) = 3.2 and f(x + 1) = 2.5f(x)?

Mathematics
1 answer:
Alex17521 [72]3 years ago
6 0
<h3>Answer:  50</h3>

========================================================

Explanation:

We're given f(1) = 3.2 which says the first term of the sequence is 3.2

To find the second term, we use the other equation, which is the recursive equation. Plug in x = 1 to get...

f(x+1) = 2.5*f(x)

f(1+1) = 2.5*f(1)

f(2) = 2.5*f(1)

f(2) = 2.5*3.2

f(2) = 8

The second term is 8. We multiplied the previous term (3.2) by the factor 2.5 to get this second term.

The third term is handled pretty much in a similar fashion

third term = 2.5*(second term)

third term = 2.5*8

third term = 20

Lastly, the fourth term f(4) is...

f(x+1) = 2.5*f(x)

f(3+1) = 2.5*f(3)

f(4) = 2.5*f(3)

f(4) = 2.5*20

f(4) = 50

The fourth term is 50.

The first four terms are: 3.2, 8, 20, 50

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Step-by-step explanation:

-3\left[\cos \left(\frac{-\pi }{4}\right)+i\sin \left(\frac{-\pi \:}{4}\right)\right]\:\div \:2\sqrt{2}\left[\cos \left(\frac{-\pi \:\:}{2}\right)+i\sin \left(\frac{-\pi \:\:\:}{2}\right)\right]

Let's apply trivial identities here. We know that cos(-π / 4) = √2 / 2, sin(-π / 4) = - √2 / 2, cos(-π / 2) = 0, sin(-π / 2) = - 1. Let's substitute those values,

\frac{-3\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)}{2\sqrt{2}\left(0-1\right)i}

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= \frac{3\left(1-i\right)}{\sqrt{2}}÷ 2\sqrt{2}i = -3-3i ÷ 4 = -\frac{3}{4}-\frac{3}{4}i

As you can see your solution is the last option.

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