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SVETLANKA909090 [29]
3 years ago
14

Find three consecutive integers such that the largest is 3 less than twice the smallest

Mathematics
1 answer:
Nostrana [21]3 years ago
7 0

Answer:

The three consecutive integers are 5,6,7

Step-by-step explanation:

Let the three consecutive integers be represented as x,x+1,x+2

As per the requirement, largest integer is 3 less than twice the smallest.

=> x+ 2 = 2x -3

=> 2x -x = 2+3

=> x= 5

So the three integers are 5,6,7

Verification:

Largest integer = 7

Smallest integer = 5

2 * smallest integer - 3 = 2* 5 - 3 = 7

Largest integer is also 7.

So the required condition in the question is satisfied by these three integers.

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Answer:

C. \frac{f(b)-f(1)}{b-1}=20

General Formulas and Concepts:

<u>Calculus</u>

  • Mean Value Theorem (MVT) - If f is continuous on interval [a, b], then there is a c∈[a, b] such that  f'(c)=\frac{f(b)-f(a)}{b-a}
  • MVT is also Average Value

Step-by-step explanation:

<u>Step 1: Define</u>

f(x)=e^{2x}

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Interval [1, b]

<u>Step 2: Check/Identify</u>

Function [1, b] is continuous.

Derivative [1, b] is continuous.

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<u>Step 3: Mean Value Theorem</u>

  1. Substitute:                    20=\frac{ f(b)-f(1)}{b-1}
  2. Rewrite:                        \frac{ f(b)-f(1)}{b-1}=20

And we have our final answer!

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