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3 years ago

The recursive function for a sequence is given below.

Mathematics

2 answers:

masya89 [10]3 years ago

8 0

Answer:

f(5) = 3200

Step-by-step explanation:

Using the recursive formula to generate the terms, that is

f(2) = 2 × f(1) = 2 × 200 = 400

f(3) = 2 × f(2) = 2 × 400 = 800

f(4) = 2 × f(3) = 2 × 800 = 1600

f(5) = 2 × f(4) = 2 × 1600 = 3200

polet [3.4K]3 years ago

3 0

Answer: 6400

Step-by-step explanation:

Given :

f(1) = 200

f(n) = 2.f(n-1) , for n = 2 ,3 , 4 , ...

To find the 5th term

when n = 2 , the sequence becomes

f(2) = 2 .f(2-1)

f(2) = 2 f(1)

since f(1) = 200

therefore:

f(2) = 2 x 200

f(2) = 400

When n = 3

f(3) = 2f(2)

f(3) = 2 x 400

f(3) = 800

When n = 4

f(4) = 2 f(3)

f(4) = 1600

When n =5

f(5) = 2f(4)

f(5) = 3200

When n = 6

f(6) = 2f(5)

f(6) = 6400

Since n = 2 , 3 , 4 ... this means that the 5th term if f(6) , therefore ,the 5th term is 6400

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Answer

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Step-by-step explanation:

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3 years ago

Five times a number subtracted from 18 is triple the number find the number

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3 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

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3 years ago

Wath is six hundred ninety- two thousand,four

ki77a [65]

692,004 is the answer

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3 years ago

Read 2 more answers

A t shirt maker wants to open his first store.If he chooses the store on Main Street he will pay 650 in rent and will charge 32$

ch4aika [34]

Answer: 35 t shirts

Step-by-step explanation:

Let number of t shirts be x

Let profit made be y

On main street, the store costs $650,

Selling the t shirt at $32 per 1

He would make a revenue of 32x

Profit = revenue - cost accrued

y1 = 32x - 650

On Broad street, the store costs $440,

Selling the t shirt at $26 per 1

He would make a revenue of 26x

Profit = revenue - cost accrued

y2 = 26x - 440

To make same profit on either location

y1 = y2

32x - 650 =26x - 440

32x -26x = -440+650

6x = 210

x = 210/6

= 35 t shirts

5 0

3 years ago