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vfiekz [6]
3 years ago
10

Use the distributive property to simplify the expression below:

Mathematics
2 answers:
Lorico [155]3 years ago
8 0
4x + 2x + 2
= 6x + 2

So the answer is B
eduard3 years ago
5 0

Answer:

B)

Step-by-step explanation:

You would first multiply the variable and constant with the number next to the parenthesis. So you now have 4x+2x+2. Then you add up like terms to get a final answer of 6x+2.

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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
How do I find point-slope form???
Nana76 [90]
I believe the answer is: By using the point slope formula, which states that a given point (x1, y1) and a slope m, the equation of the line can be written as y - y1 = m(x - x1)
8 0
3 years ago
I need help with math
Effectus [21]
what do you need help with, maybe I can help? 

4 0
3 years ago
2 taps A and B can fill a swimming pool in 3 hours. If turned on alone, it takes tap A 5 hours less than tap B to fill the same
vazorg [7]

Answer:

Tap A will take 2 hours and tap B will take 6 hours to fill the tank when turned on alone.

Step-by-step explanation:

Let tap B fills the pool alone in the time = x hours

So in one hour part of pool will be filled = \frac{1}{x}

Another tap A when turned on, it takes time to fill the pool = x-5 hours

So in one hour part of the same pool filled = \frac{1}{x-5}

Now both the taps A and B are turned on then time taken to fill the pool = 3 hours.

Part of the pool filled in one hour by both the taps = \frac{1}{3}

Now we form an equation

Part of pool filled in one hour by tap A + Part of pool filled in one hour by tap B = Part of pool filled in one hour by both the taps when turned on

\frac{1}{(x-5)}+\frac{1}{x}=\frac{1}{3}

\frac{x-5+1}{x(x-5)}=\frac{1}{3}

\frac{x-4}{x(x-5)}=\frac{1}{3}

3(x - 4) = x(x - 5)

x² -5x = 3x - 12

x² - 8x + 12 = 0

x² - 6x - 2x + 12 = 0

x(x - 6) - 2(x - 6) = 0

(x -2)(x - 6) = 0

x = 2, 6 hours

We will take higher value of x as x = 6 hours for tap B.

Time taken by tap A = 6 - 4 = 2 hours.

Therefore, Tap A will take 2 hours and tap B will take 6 hours to fill the tank when turned on alone.

4 0
3 years ago
How do you translate the equation 4x-5=3(x+2) into a verbal sentence
elena55 [62]
Four times the quantity of x minus five is equal to three times the quantity of x plus two
3 0
3 years ago
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