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3 years ago

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If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possi

ble value of n?
A. 10B. 11C. 12D. 13E. 14

1 answer:

blagie [28]3 years ago

3 0

Answer:

B. 11

Step-by-step explanation:

When n = 11

The product of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11 is 39,916,800

39,916,800 is a multiple of 990

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Evgesh-ka [11]

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3 years ago

A manufacturing company with 450 employees begins a new product line and must increase their numbers of employees by 18%. How ma

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You are adding 18%, which makes it 118% of original employees

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2 years ago

Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction

Oliga [24]

The base case is the claim that

$\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}$

which reduces to

$\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86$

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}$

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}$

By the induction hypothesis,

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}$

Now compare this to the upper bound we seek:

$\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}$

because

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2 years ago

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MatroZZZ [7]

6
6 * 0.7 = 4.2
4.2 * 0.7 = 2.94
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4.2 + 4.2 + 2.94 + 2.94 + 2.058 + 2.058 + 1.4406 + 1.4406 =

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2 years ago

Giving 20 points for the answer.

solong [7]

Answer:

B, A, C

Step-by-step explanation:

add the like terms together (ending in x2 with x2, ending in x with x etc )

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1 year ago

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