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1 year ago

The number 2.2360667… is a __ because it cannot be written as a fraction(ratio)

1 answer:

8 0

in our number " 2.2360667…" the 3 dots mean that the digits keep going forever, so we conclude that our number belongs to the set of irrational numbers.

<h3>Which type of number is 2.2360667…?</h3>

We will define two types of numbers:

Rational numbers: Are these that can be written as the quotient of two integers.

Irrational numbers: Can't be written as the quotient of two integer numbers, A rational number always has an infinite number of digits after the decimal point, and there is no pattern in these digits.

Now, in our number " 2.2360667…" the 3 dots mean that the digits keep going forever, that is enough to conclude that our number belongs to the set of irrational numbers.

If you want to learn more about rational numbers:

brainly.com/question/12088221

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What is the product (-6a^3b+2ab^2

Irina-Kira [14]

you have to apply this formula (a+b)(c-d)=ac-ad+bc-bd

-6a^3*b*5a^2+6a^3b*2ab+6a^3b*b+2ab^2*5a^2-2ab^2*2ab-2ab^2*b=

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2 years ago

consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where

boyakko [2]

Answer:

a) $X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

b) $Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

Step-by-step explanation:

From the question we are told that

The Function

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

Generally the differentiation of function f(x) is mathematically solved as

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

$f(x)=\frac{x^3+x^2+5x+1}{x^2}$

Therefore

$f'(x)=\frac{x^2+10x+3}{x^4}$

Generally critical point is given as

$f'(x)=0$

$\frac{x^2+10x+3}{x^4}=0$

$x=-5 \pm\sqrt{22}$

Generally the maximum and minimum x value for critical point is mathematically solved as

$f'(-5 \pm\sqrt{22})$

Where

Maximum value of x

$f'(-5 +\sqrt{22})$

Minimum value of x

$f'(-5 +\sqrt{22})$

Therefore interval of increase is mathematically given by

$f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})$

$f(x)$

Therefore interval of decrease is mathematically given by

$(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)$

Generally the second differentiation of function f(x) is mathematically solved as

$f''(x)=\frac{2(x^2+15x+6)}{x^5}$

Generally the point of inflection is mathematically solved as

$f''(x)=0$

$x^2+15x+6=0$

Therefore inflection points is given as

$x=\frac{1}{2} (-15 \pm \sqrt{201}$

$f''(x)>0,\frac{1}{2}(-15-\sqrt{201})$

a)Generally the concave upward interval X is mathematically given as

$X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

$f''(x)$

b)Generally the concave downward interval Y is mathematically given as

$Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

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