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MaRussiya [10]
3 years ago
13

If a human face only have 2 eye balls how many eye balls can rap there whole head? 1-10 in why?

Mathematics
1 answer:
IceJOKER [234]3 years ago
7 0
22 I hope this helps
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HOW DO THE EXPRESSION 4(3+X) AND (3+3+3+3)+(X+X+X+X) COMPARE TO ANOTHER
UkoKoshka [18]
They are equal to one another because 4(3+x) can be rewritten using the distributive property as 4(3)+4x or 3+3+3+3+x+x+x+x
8 0
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Analyze ady Porter is buying t ride tickets at the county fair. He receives 3 tickets for every dollar, d, that he spends. Which
Vanyuwa [196]

Answer:

The answer is below

Step-by-step explanation:

An independent variable is a variable that does not dependent on other variables. It is the input variable.

A dependent variable is a variable that is dependent on other variables. The dependent variable depends on the independent variable. It is the output variable.

Since porter receives 3 tickets for every dollar, this means the number of ticket is dependent on the money he spend. Therefore the money he spends (d) is the independent variable and the tickets (t) is the dependent variable.

4 0
2 years ago
What is the answer of 11 6/9 - 4 4/9
IrinaK [193]

Answer:

7  \frac{2}{9}

Step-by-step explanation:

here is your answer

Calculation steps:

11  \frac{6}{9}   - 4 \frac{4}{9}

= (14 - 4)  + ( \frac{6}{9}  -  \frac{4}{9} )

= 7 +  \frac{6 - 4}{9}

=  7 +  \frac{2}{9}

7  \frac{2}{9}

4 0
2 years ago
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A line has a slope of \large \frac{5}{3}. Which of the following points could this line pass through?
kiruha [24]

Answer:well I will tell if you tell me the answer for my question

Step-by-step explanation:

6 0
3 years ago
Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre
Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = \frac{d}{dx}[x^{4}ln(x)]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = 4x^{3}ln(x) + x_{4}.\frac{1}{x}

f'(x) = 4x^{3}ln(x) + x^{3}

f'(x) = x^{3}[4ln(x) + 1]

Now, find the critical points: f'(x) = 0

x^{3}[4ln(x) + 1] = 0

x^{3} = 0

x = 0

and

4ln(x) + 1 = 0

ln(x) = \frac{-1}{4}

x = e^{\frac{-1}{4} }

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval                 x-value                      f'(x)                       result

0<x<0.78                 0.5                 f'(0.5) = -0.22            decreasing

x>0.78                       1                         f'(1) = 1                  increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

  • Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
  • After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = x^{4}ln(x)

f(0.78) = 0.78^{4}ln(0.78)

f(0.78) = - 0.092

The point of <u>minimum</u> is (0.78, - 0.092)

(c) To determine the inflection point (IP), calculate the second derivative of the function and solve for x:

f"(x) = \frac{d^{2}}{dx^{2}} [x^{3}[4ln(x) + 1]]

f"(x) = 3x^{2}[4ln(x) + 1] + 4x^{2}

f"(x) = x^{2}[12ln(x) + 7]

x^{2}[12ln(x) + 7] = 0

x^{2} = 0\\x = 0

and

12ln(x) + 7 = 0\\ln(x) = \frac{-7}{12} \\x = e^{\frac{-7}{12} }\\x = 0.56

Substituing x in the function:

f(x) = x^{4}ln(x)

f(0.56) = 0.56^{4} ln(0.56)

f(0.56) = - 0.06

The <u>inflection point</u> will be: (0.56, - 0.06)

In a function, the concave is down when f"(x) < 0 and up when f"(x) > 0, adn knowing that the critical points for that derivative are 0 and 0.56:

f"(x) =  x^{2}[12ln(x) + 7]

f"(0.1) = 0.1^{2}[12ln(0.1)+7]

f"(0.1) = - 0.21, i.e. <u>Concave</u> is <u>DOWN.</u>

f"(0.7) = 0.7^{2}[12ln(0.7)+7]

f"(0.7) = + 1.33, i.e. <u>Concave</u> is <u>UP.</u>

4 0
3 years ago
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