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3 years ago

15

Apply the distributive property to factor out the greatest common factor 12+80

1 answer:

Nikolay [14]3 years ago

5 0

Answer:

<em>4</em>

Step-by-step explanation:

<em>First, write out the prime factorization of both 12 and 80: 12 = 2 x 2 x 3 and 80 = 2 x 2 x 2 x 2 x 5. Now collect only the prime numbers that each of them share in their factorizations, which are only two 2's. Thus, the greatest common factor between 12 and 80 is 2 x 2 = 4.</em>

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What does 2*-2-(3- -2) equal

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Step-by-step explanation:

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Find a positive number for which the sum of it and its reciprocal is the smallest (least) possible. Let x be the number and let

Allisa [31]

Answer:

$S(x) = x + \frac{1}{x}$ --- Objective function

Interval = $\{x:x=1\}$

Step-by-step explanation:

Given

Represent the number with x

The required sum can be represented as:

$x + \frac{1}{x}$

Hence, the objective function is:

$S(x) = x + \frac{1}{x}$

To get the the interval, we start by differentiating w.r.t x

<em>Using first principle, this gives:</em>

$S'(x) = 1 - \frac{1}{x^2}$

Equate S'(x) to 0 in order to solve for x

$0 = 1 - \frac{1}{x^2}$

Subtract 1 from both sides

$0 -1 = 1 -1 - \frac{1}{x^2}$

$-1 = - \frac{1}{x^2}$

Multiply both sides by -1

$1 = \frac{1}{x^2}$

Cross Multiply

$x^2 * 1 = 1$

$x^2 = 1$

Take positive square root of both sides because x is positive

$\sqrt{x^2} = \sqrt{1$

$x = 1$

Representing x using interval notation, we have

Interval = $\{x:x=1\}$

To get the smallest sum, we substitute 1 for x in $S(x) = x + \frac{1}{x}$

$S(1) = 1 + \frac{1}{1}$

$S(1) = 1 + 1$

$S(1) = 2$

<em>Hence, the smallest sum is 2</em>

3 0

3 years ago