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devlian [24]
2 years ago
8

Consider the first three terms of the sequence below.

Mathematics
1 answer:
Simora [160]2 years ago
5 0

Answer:

f(1) = <u> 14000 </u>

f(n) = f(n - 1) · <u> 0.9 </u>, for n ≥ 2

The next term in the sequence is <u> 10206 </u>.

Step-by-step explanation:

First we try to find the type of sequence it is.

Is there a common difference?

12600 - 14000 = -1400

11340 - 12600 = -1260

There is no common difference, so it is not an arithmetic sequence.

12600/14000 = 0.9

11340/12600 = 0.9

There is a common ratio, so this is a geometric sequence.

Each term is 0.9 times the previous term.

The next term is:

11340 * 0.9 = 10206

Answer:

f(1) = <u> 14000 </u>

f(n) = f(n - 1) · <u> 0.9 </u>, for n ≥ 2

The next term in the sequence is <u> 10206 </u>.

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<h3>How to solve for x?</h3>

The complete question is in the attached image.

From the attached image of the triangle, we can see that the triangle is a right triangle, and x can be solved using the following sine function

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Solve for x

x = \frac{7\sqrt 2}{\sqrt 2}

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x = 7

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Given: <br> PQ<br> ⊥<br> QR<br> , PR=20,<br> SR=11, QS=5<br> Find: The value of PS.
dlinn [17]

Answer:

The value of the side PS is 26 approx.

Step-by-step explanation:

In this question we have two right triangles. Triangle PQR and Triangle PQS.

Where S is some point on the line segment QR.

Given:

PR = 20

SR = 11

QS = 5

We know that QR = QS + SR

QR = 11 + 5

QR = 16

Now triangle PQR has one unknown side PQ which in its base.

Finding PQ:

Using Pythagoras theorem for the right angled triangle PQR.

PR² = PQ² + QR²

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PQ = √656

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Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.

Finding PS:

Using Pythagoras theorem, we have:

PS² = PQ² + QS²

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