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Ipatiy [6.2K]
2 years ago
8

chose each. subset of the real number system that can be used to classify the number 5.33 rational irrational number integer who

le number natural number
Mathematics
1 answer:
tresset_1 [31]2 years ago
8 0

The subset of real number that can be used to classify the number 5.33 is irrational number.

<h3 /><h3>Classifying the number</h3>

We know that 5.33 = 5 + 0.33 contains a decimal part which is 0.33 and 5 which is a whole number part.

Since 0.33 is the irrational or decimal part which makes 5 + 0.33 = 5.33 irrational since a rational number + an irrational number = irrational number.

Since an irrational number is a number which cannot be expressed as a ratio of two integers.

<h3 /><h3>Conclusion</h3>

So, 5.33 is an irrational number.

So, the subset of real number that can be used to classify the number 5.33 is irrational number.

Learn more about irrational number here:

brainly.com/question/5342576

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Answer:

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Height is 2

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Volume = 3.14 × 9 × 2

Volume = 3.14 × 18

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<span>if

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African elephants can eat up to 500 pounds of vegetation in one day. How many pounds of vegetation could an elephant eat in a we
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Step-by-step explanation:

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2 years ago
Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter
Vaselesa [24]

We have been given a function f(x)=2x^3+3x^2-336x. We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

f'(x)=2(3)x^{2}+3(2)x^1-336

f'(x)=6x^{2}+6x-336

6x^{2}+6x-336=0

x^{2}+x-56=0

x^{2}+8x-7x-56=0

(x+8)-7(x+8)=0

(x+8)(x-7)=0

x=-8,x=7

So x=-8,7 are critical points and these will divide our function in 3 intervals (-\infty,-8)U(-8,7)U(7,\infty).

Now we will find derivative over each interval as:

f'(x)=(x+8)(x-7)

f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16

Since f'(9)>0, therefore, function is increasing on interval (-\infty,-8).

f'(x)=(x+8)(x-7)

f'(1)=(1+8)(1-7)=(9)(-6)=-54

Since f'(1), therefore, function is decreasing on interval (-8,7).

Let us check for the derivative at x=8.

f'(x)=(x+8)(x-7)

f'(8)=(8+8)(8-7)=(16)(1)=16

Since f'(8)>0, therefore, function is increasing on interval (7,\infty).

(b) Since x=-8,7 are critical points, so these will be either a maximum or minimum.

Let us find values of f(x) on these two points.

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(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

f''(x)=6(2)x^{1}+6

f''(x)=12x+6

12x+6=0

12x=-6

\frac{12x}{12}=-\frac{6}{12}

x=-\frac{1}{2}

Therefore, x=-\frac{1}{2} is an inflection point of given function.

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Answer:

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