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statuscvo [17]
3 years ago
6

What is the equivalent recursive definition for an = 12+ (n - 1)3?

Mathematics
1 answer:
Verdich [7]3 years ago
6 0

Answer:

A_1 = 12

A_n = A_{n-1} + 3

Step-by-step explanation:

Given

A_n =12+(n-1)3

Required

Write as recursive

We have:

A_n =12+(n-1)3

Open bracket

A_n =12+3n-3

A_n =12-3+3n

A_n =9+3n

Calculate few terms

A_1 =9+3*1 = 9 + 3 = 12

A_2 =9+3*2 = 9 + 6 = 15

A_3 =9+3*3 = 9 + 9 = 18

The above shows that the rule is to add 3.

So, we have:

A_1 = 12

A_n = A_{n-1} + 3

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By flipping the second fraction (finding its reciprocal) we change the equation mathematically. In order to maintain the equation mathematically, we must turn the division question into a multiplication question.

Step-by-step explanation:

4 0
3 years ago
Need help with this.
stepan [7]
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6 0
3 years ago
Please help, need an answer ASAP!
Shtirlitz [24]

Answer:

6, -7

Step-by-step explanation:

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6 0
3 years ago
50 points and BRAINLIEST<br><br> .15m - .13m = 69.96 - 55.96
weeeeeb [17]

Answer:

m=7

Step-by-step explanation:

15m−13m=69.96−55.96

Step 1: Simplify both sides of the equation.

15m−13m=69.96−55.96

15m+−13m=69.96+−55.96

(15m+−13m)=(69.96+−55.96)(Combine Like Terms)

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2m=14

Step 2: Divide both sides by 2.

2m

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=

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5 0
3 years ago
Suppose b is any integer. If b mod 12 = 5, then 7b mod 12
Rzqust [24]

The result follows directly from properties of modular arithmetic:

b\equiv5\pmod{12}\implies 7b\equiv35\equiv-1\equiv\boxed{11}\pmod{12}

That is,

b\equiv5\pmod{12}

means we can write b=12n+5 for some integer n. Then

7b=7(12n+5)=12(7n)+35

and taken mod 12, the first term goes away, so

7b\equiv35\pmod{12}

etc

8 0
3 years ago
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