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Levart [38]
2 years ago
9

5. What is the second step in proving by mathematical induction that for every positive integer n, 11" - 6 is divisible by

Mathematics
2 answers:
WARRIOR [948]2 years ago
4 0

Answer:

Step-by-step explanation:

Premise:  If n is a positive integer, then 11ⁿ-6 is divisible by 5.

Let n=1.

11¹-6 = 5, so the premise is true for n=1.

Suppose the premise is true when n is an integer greater than 1,

then 11ⁿ-6 is divisible by 5.

11ⁿ⁺¹-6 = 11·11ⁿ - 6

   = (10+1)11ⁿ - 6

   = (10·11ⁿ) + (1·11ⁿ) - 6

   = (10·11ⁿ) + (11ⁿ - 6)

Both terms are divisible by 5, so 11ⁿ⁺¹-6 is divisible by 5. Therefore, the premise holds true for n+1.

Proof by induction.

astraxan [27]2 years ago
3 0

Answer:

The correct answer is A:Let n = k. Then assume that 11" - 6 is divisible by 5. sorry if im wrong but if correct pls mark brainliest

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Plz help and show all work lots of brainly points :D
Lunna [17]

Answer:

the answer is A

Step-by-step explanation:

8 0
3 years ago
List a value of b that will cause 4x2 + bx + 25 = 0 to have one real solution
vlada-n [284]

Answer:

20

Step-by-step explanation:

If you take (2x+5) (2x+5) = 0

then 4x^2 + 20x +25 = 0

5 0
3 years ago
Help !!! 8th grade math
Ipatiy [6.2K]
#2=4
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3 years ago
Read 2 more answers
Two candidates are randomly selected. a. List all the possible outcomes. b. What is the probability that both are qualified? c.
steposvetlana [31]

Question:

Four candidates are to be interviewed for a job. Two of them, numbered 1 and 2, are qualified, and the other two, numbered 3 and 4, are not.

Answer:

a. S = {12,13,14,23,24,34}

b. 1/6

c. ⅔

Step-by-step explanation:

Let S = Sample Space i.e. total possible outcomes

Given that candidates 1 and 2 are qualified and 3 and 4 are not qualified.

If two candidates are randomly selected. The list of all possible outcomes is as follows

S:{12,13,14,23,24,34}

And the number of outcomes is 6

b. What is the probability that both are qualified?

The event that both are qualified is given as {12} or {21}

From the sample space in (a) above, there's only one occurrence of {12} = 1

So, the probability is calculated as 1/6

c. What is the probability that exactly one is qualified?

The event that one one is qualified is given as {13,14,23,24}

Total = 4

Probability = 4/6

Probability = ⅔

5 0
3 years ago
In order to get to her Grandmother's house, Little Red Riding Hood walks 6 miles
Vesnalui [34]

Answer:

4.4 miles

Step-by-step explanation:

c= sqrt 6^2+10^2

sqrt 36+100

Sqrt 136=11.66190378

10+6=16

16-11.66190378=4.33809622

3 0
3 years ago
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