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horsena [70]
3 years ago
15

;)

Mathematics
2 answers:
maksim [4K]3 years ago
4 0

Answer:

Not really a math question but I'll answer anyway (I don't really listen to Doja Cat, so I don't have one)

Step-by-step explanation:

Ariana Grande: Positions

Selena Gomez: Wolves

Have a good day :)

mafiozo [28]3 years ago
3 0

Answer:

Side to side

I love all of hers

And streets

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Aneli [31]

Answer:

B. 36

Step-by-step explanation:

The answer is B. 36.

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4 years ago
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Can you determine that ABC = DCB ? justify your answer
Free_Kalibri [48]

Answer:

You cant.

Step-by-step explanation:You woud need to know the variables.They have to be the same but can be in a diffrent order

3 0
3 years ago
Which number is irrational? A. 3 over 8 B. square root of 56 over 8 end root C. fraction numerator square root of 36 over denomi
Rasek [7]
The correct answer is letter <span> D. Syntax error from line 1 column 53 to line 1 column 60. Unexpected '0'. In order for a number to be considered as irrational when there is no possible ratio for that number. Among the four choices, choice D only tells us that the possible ratio or fraction has a denominator of Zero.</span>
7 0
3 years ago
What is absolute extrema of cube root of x on I=[-3,8]
hichkok12 [17]
<span>These are points where f ' = 0. Use the quiotent rule to find f '. 

f ' (x) = [(x^3+2)(1) - (x)(3x^2)] / (x^3+2)^2 
f ' (x) = (2 - 2x^3) / (x^3 + 2)^2 

Set f ' (x) = 0 and solve for x. 

f ' (x) = 0 = (2-2x^3) / (x^3+2)^2 

Multiply both sides by (x^3+2)^2 

(x^3+2)^2 * 0 = (x^3+2)^2 * [(2-2x^3)/(x^3+2)^2] 
0 = 2 - 2x^3 

Add 2x^3 to both sides 

2x^3 + 0 = 2x^3 + 2 - 2x^3 
2x^3 = 2 

Divide both sides by 2 

2x^3 / 2 = 2 / 2 
x^3 = 1 

Take cube roots of both sides 

cube root (x^3) = cube root (1) 
x = 1. This is our critical point 

2) Points where f ' does not exist. 

We know f ' (x) = (2-2x^3) / (x^3+2)^2 

You cannot divide by 0 ever so f ' does not exist where the denominator equals 0 

(x^3 + 2)^2 = 0. Take square roots of both sides 
sqrt((x^3+2)^2) = sqrt(0) 
x^3 + 2 = 0. Add -2 to both sides. 
-2 + x^3 + 2 = -2 + 0 
x^3 = -2. Take cube roots of both sides. 
cube root (x^3) = cube root (-2) 
x = cube root (-2). This is where f ' doesnt exist. However, it is not in our interval [0,2]. Thus, we can ignore this point. 

3) End points of the domain. 

The domain was clearly stated as [0, 2]. The end points are 0 and 2. 

Therefore, our only options are: 0, 1, 2. 

Check the intervals 

[0, 1] and [1, 2]. Pick an x value in each interval and determine its sign. 

In [0, 1]. Check 1/2. f ' (1/2) = (7/4) / (17/8)^2 which is positive. 

In [1, 2]. Check 3/2. f ' (3/2) = (-19/4) / (43/8)^2 which is negative. 

Therefore, f is increasing on [0, 1] and decreasing on [1, 2] and 1 is a local maximum. 

f (0) = 0 
f (1) = 1/3 
f (2) = 1/5 

Therefore, 0 is a local and absoulte minimum. 1 is a local and absolute
maximum. Finally, 2 is a local minimum. </span><span>Thunderclan89</span>
3 0
4 years ago
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bixtya [17]

Answer: 7.1

Step-by-step explanation:

7.09<7.10

7.10=7.1

sometimes adding a zero helps

3 0
2 years ago
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