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denis-greek [22]
2 years ago
11

What is the fifth term of the sequence defined by the recursive rule f(1) = 5, f(n)=f(n-1) + 3

Mathematics
1 answer:
Oxana [17]2 years ago
7 0

Answer:

f(5) = 17

Step-by-step explanation:

Using the recursive rule and f(1) = 5 , then

f(2) = f(1) + 3 = 5 + 3 = 8

f(3) = f(2) + 3 = 8 + 3 = 11

f(4) = f(3) + 3 = 11 + 3 = 14

f(5) = f(4) + 3 = 14 + 3 = 17

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Factorise 12x - 30<br> help me please
Sonbull [250]

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Step-by-step explanation:

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2 years ago
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The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis
Firdavs [7]

Answer:

sin\theta_1 =  - \frac{2\sqrt{70}}{19}

Step-by-step explanation:

We are given that \theta_1 is in <em>fourth</em> quadrant.

cos\theta_1 is always positive in 4th quadrant and  

sin\theta_1 is always negative in 4th quadrant.

Also, we know the following identity about sin\theta and cos\theta:

sin^2\theta + cos^2\theta = 1

Using \theta_1 in place of \theta:

sin^2\theta_1 + cos^2\theta_1 = 1

We are given that cos\theta_1 = \frac{9}{19}

\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 =  \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 =  \dfrac{280}{361}\\\Rightarrow sin\theta_1 =  \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 =  +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}

\theta_1 is in <em>4th quadrant </em>so sin\theta_1 is negative.

So, value of sin\theta_1 =  - \frac{2\sqrt{70}}{19}

6 0
2 years ago
Find the sum or difference. a. -121 2 + 41 2 b. -0.35 - (-0.25)
s344n2d4d5 [400]

Answer:

2

Step-by-step explanation:

The reason an infinite sum like 1 + 1/2 + 1/4 + · · · can have a definite value is that one is really looking at the sequence of numbers

1

1 + 1/2 = 3/2

1 + 1/2 + 1/4 = 7/4

1 + 1/2 + 1/4 + 1/8 = 15/8

etc.,

and this sequence of numbers (1, 3/2, 7/4, 15/8, . . . ) is converging to a limit. It is this limit which we call the "value" of the infinite sum.

How do we find this value?

If we assume it exists and just want to find what it is, let's call it S. Now

S = 1 + 1/2 + 1/4 + 1/8 + · · ·

so, if we multiply it by 1/2, we get

(1/2) S = 1/2 + 1/4 + 1/8 + 1/16 + · · ·

Now, if we subtract the second equation from the first, the 1/2, 1/4, 1/8, etc. all cancel, and we get S - (1/2)S = 1 which means S/2 = 1 and so S = 2.

This same technique can be used to find the sum of any "geometric series", that it, a series where each term is some number r times the previous term. If the first term is a, then the series is

S = a + a r + a r^2 + a r^3 + · · ·

so, multiplying both sides by r,

r S = a r + a r^2 + a r^3 + a r^4 + · · ·

and, subtracting the second equation from the first, you get S - r S = a which you can solve to get S = a/(1-r). Your example was the case a = 1, r = 1/2.

In using this technique, we have assumed that the infinite sum exists, then found the value. But we can also use it to tell whether the sum exists or not: if you look at the finite sum

S = a + a r + a r^2 + a r^3 + · · · + a r^n

then multiply by r to get

rS = a r + a r^2 + a r^3 + a r^4 + · · · + a r^(n+1)

and subtract the second from the first, the terms a r, a r^2, . . . , a r^n all cancel and you are left with S - r S = a - a r^(n+1), so

(IMAGE)

As long as |r| < 1, the term r^(n+1) will go to zero as n goes to infinity, so the finite sum S will approach a / (1-r) as n goes to infinity. Thus the value of the infinite sum is a / (1-r), and this also proves that the infinite sum exists, as long as |r| < 1.

In your example, the finite sums were

1 = 2 - 1/1

3/2 = 2 - 1/2

7/4 = 2 - 1/4

15/8 = 2 - 1/8

and so on; the nth finite sum is 2 - 1/2^n. This converges to 2 as n goes to infinity, so 2 is the value of the infinite sum.

8 0
3 years ago
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