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castortr0y [4]
3 years ago
9

Calculating Profit Using this table, calculate the profit at each level of bicycle production. One bike: -$ Two bikes: $ Three b

ikes: $ Four bikes: $ Five bikes: $ Six bikes: $ Seven bikes: $
Mathematics
1 answer:
leonid [27]3 years ago
8 0

Answer:

1~30

2~3

3~40

4~70

5~90

6~90

7~80

Step-by-step explanation:

subtract the total revenue from the total cost

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Solve: |x+2|−1≥5. Write your solution in interval notation.
AlekseyPX

nswer:

Step-by-step explanat

|x+2| - 1 ≥ 5

  |x+2|  ≥ 5+1

 |x+2|  ≥ 6

 x+2 ≥ 6 hoặc x+2 ≤ -6

+ với x+2 ≥ 6  x ≥ 6 – 2   x ≥ 4

+ với x+2 ≤ -6  x ≤ -6 – 2  x ≤ -8

(-∝;-8)∪(4;+∝)

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
Stephany is inviting 3 friends to a party. If each friend will get 7 cookies and each box has 10 cookies, how many boxes does St
Aleksandr [31]

Answer:

3 boxex hope that helps. yea. thats the answer

7 0
2 years ago
James found a pair of shoes he wanted at two stores. At store A they are $35 and he had a 20% off coupon. At store B they were $
Oxana [17]

Answer: They cost the same amount.

Step-by-step explanation:

35(0.20)=7

35-7=28

40(0.30)=12

40-12=28

28=28

7 0
3 years ago
Give a correct interpretation of a 99% confidence level of 0.102 &lt; p &lt; 0.236.
lord [1]
We are 99% confident that the interval from 0.102 to 0.236 actually does contain the true value of the population proportion <span>p.</span>
4 0
3 years ago
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