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lbvjy [14]
3 years ago
13

Please someone help me, i need thier solve please my teacher told me to solve them ​

Mathematics
1 answer:
mariarad [96]3 years ago
6 0

Answer:

A. Domain : (-∞, ∞)

B. Function is increasing in the interval (-2, 0) and (2, ∞)

   Decreasing in the interval of (-∞, -2) and (0, 2).

Step-by-step explanation:

A. Given function is, y = |2x - 1|

This function is the transformed form of the parent function, y = |x|

Domain of the parent function is { x | x is a set of real numbers}

Therefore, domain of the transformed function will be the same as the domain of the parent function.

Domain of the function = {x | x is a real number}

B. Given function is f(x) = (x^2-4) ^{\frac{2}{3} }

Domain of the function : (-∞, ∞)

Critical points of the function are,

⇒ x = 0, ±2

Now we find the three intervals where we have to check the function to be increasing or decreasing.

(-∞ -2), (-2, 0), (0, 2), (2, ∞)

Derivative of the function f(x),

f'(x) = \frac{4x}{3(x^2-4)^{\frac{1}{3} } }

Here, f'(x) < 0 for (-∞, -2)

f'(x) > 0 for (-2, 0)

f'(x) < 0 for (0, 2)

f'(x) > 0 for (2, ∞)

Therefore, given function is increasing in the interval (-2, 0) and (2, ∞)

And it's decreasing in the interval of (-∞, -2), and (0, 2).

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What is 65 divided by 7,570
Ostrovityanka [42]

Answer:

7,570/65

Step-by-step explanation:

4,920.5

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65/7570

0.0085865258

5 0
3 years ago
If k stands for an integer, then is it possible for k2 + k to stand for an odd integer? Be prepared to justify your answer.
BartSMP [9]

Answer:

<em>k^2 + k never stands for an odd integer</em>

Step-by-step explanation:

Let us consider either case with which k stands for an odd or even integer;

Case 1: k is an odd integer

For integer a, k = 2a + 1

So, k + 1 = 2a + 2 = 2( a + 1 ) = 2b for integer b

k^2 + k = k ( k + 1 ) = k ( 2b ) = 2kb = 2c for integer c,

<em>Therefore, if k is an odd integer, then k^2 + k is an even integer ;</em>

Case 2: k is an even integer

For an integer a, k = 2a

So, k + 1 = 2a + 1

k^2 + k = k( k+1 ) = 2a( 2a + 1 ) , multiple of 2

<em>Therefore, if k is an even integer, then k^2 + k is an even integer;</em>

<em>This would make k^2 + k never stand for an odd integer</em>

5 0
3 years ago
Find the circumference of a circle that has an area of 452.16 square meters. (Use 3.1416 as the value of π.)
boyakko [2]
452.16 = 3.1416 r^2
r^2 = 452.16  /  3.1416
r^2 = 143.93
r = 11.997

C = 2 (3.1416) (12) =75.38 m
3 0
4 years ago
You purchased a stock at a price of $51.17. The stock paid a dividend of $1.75 per share and the stock price at the end of the y
ipn [44]

Answer: i think it is 109.51

Step-by-step explanation:

7 0
3 years ago
Check
almond37 [142]

Answer:

ayo

Step-by-step explanation:

6 0
3 years ago
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