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Y_Kistochka [10]
3 years ago
5

10times10times10times10times10times10

Mathematics
1 answer:
BARSIC [14]3 years ago
7 0

Answer:

1000000

Step-by-step explanation:

10x10x10x10x10x10=1000000

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H + 169 = 759 solve for h
steposvetlana [31]

Answer:

h = 590

Step-by-step explanation:

To find x, rearrange the equation to where h is on its own side.

<u>Rearranged equation:</u>

759 - 169 = h

Since 169 added to h was to find 769, 169 subtracted from 759 represents h.

<u>Solve:</u>

759 - 169 = 590

590 = h.

5 0
2 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
The girls soccer team won 10 games and the lost 2. And the boys soccer team won 12 games and lost 3. What is the ratio of the gi
worty [1.4K]
5 to 1 because they lost 10 games in the time they lost 2 games so it would in smallest form at 5 to 1
4 0
3 years ago
Don’t understand please help
Mazyrski [523]
A is ur answer trust me
7 0
3 years ago
What is a possible value of sin(x) if the cos(2x) = 0.42
Ostrovityanka [42]

Answer:

The answer is 0.54 have a great day

6 0
3 years ago
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