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LuckyWell [14K]
3 years ago
14

Please help on number 15

Mathematics
1 answer:
malfutka [58]3 years ago
3 0
76 because what u do to one number is done to the other so u have to subtract 24 to both numbers.    Can u mark me as brainliest??
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1) Let f(x)=6x+6/x. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relat
brilliants [131]

Answer:

1) increasing on (-∞,-1] ∪ [1,∞), decreasing on [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum

2) increasing on [1,∞), decreasing on (-∞,0) ∪ (0,1]

x = 1 is absolute minimum

3) increasing on (-∞,0] ∪ [8,∞), decreasing on [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum

4) increasing on [2,∞), decreasing on (-∞,2]

x = 2 is absolute minimum

5) increasing on the interval (0,4/9], decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum

Step-by-step explanation:

To find minima and maxima the of the function, we must take the derivative and equalize it to zero to find the roots.

1) f(x) = 6x + 6/x

f\prime(x) = 6 - 6/x^2 = 0 and x \neq 0

So, the roots are x = -1 and x = 1

The function is increasing on the interval (-∞,-1] ∪ [1,∞)

The function is decreasing on the interval [-1,0) ∪ (0,1]

x = -1 is local maximum, x = 1 is local minimum.

2) f(x)=6-4/x+2/x^2

f\prime(x)=4/x^2-4/x^3=0 and x \neq 0

So the root is x = 1

The function is increasing on the interval [1,∞)

The function is decreasing on the interval (-∞,0) ∪ (0,1]

x = 1 is absolute minimum.

3) f(x) = 8x^2/(x-4)

f\prime(x) = (8x^2-64x)/(x-4)^2=0 and x \neq 4

So the roots are x = 0 and x = 8

The function is increasing on the interval (-∞,0] ∪ [8,∞)

The function is decreasing on the interval [0,4) ∪ (4,8]

x = 0 is local maximum, x = 8 is local minimum.

4) f(x)=6(x-2)^{2/3} +4=0

f\prime(x) = 4/(x-2)^{1/3} has no solution and x = 2 is crtitical point.

The function is increasing on the interval [2,∞)

The function is decreasing on the interval (-∞,2]

x = 2 is absolute minimum.

5) f(x)=8\sqrt x - 6x for x>0

f\prime(x) = (4/\sqrt x)-6 = 0

So the root is x = 4/9

The function is increasing on the interval (0,4/9]

The function is decreasing on the interval [4/9,∞)

x = 0 is local minimum, x = 4/9 is absolute maximum.

5 0
2 years ago
Which of the following is not approximately equivalent to one of the metric units: 1 kilometer, 1 meter, 1 kilogram, or 1 liter?
Lubov Fominskaja [6]
Liter is not equivalent to one of the metric units
3 0
3 years ago
Please help im struggling and it’s due in 30 minutes
MAXImum [283]
I know this is not an answer but if your having trouble with this there is a website called Desmos all you do is type in the equation for the graph and it gives you the answer
6 0
3 years ago
Is (4,-2) (5,-3), (6,-2), (7,-3) a function??? need answers quick pls(ᗒᗣᗕ)՞
erastova [34]

Answer:

7.-3

Step-by-step explanation:

yes.

7 0
2 years ago
Avicenna, a major insurance company, offers five-year life insurance policies to 65-year-olds. If the holder of one of these pol
ki77a [65]

Answer:

Avicenna can expect to lose money from offering these policies. In the long run, they should expect to lose ___33__ dollars on each policy sold

Step-by-step explanation:

Given :

The amount the company Avicenna must pay to the shareholder if the person die before 70 years = $ 26,500

The value of each policy = $497

It is given that there is a 2% chance that people will die before 70 years and 98% chance that people will live till the age 70.

The expected policy to be sold= policy nominal + chances of death

                                      = 497 + [98% (no pay) + 2% (pay)]

                                     = 497 + [98%(0) + 2%(-26500)]

(The negative sign shows that money goes out of the company)

                                   = 497 - 2% (26500)

                                  = 497 - 530

                                  =33

Therefore the company loses 33 dollar on each policy sold in the long run.

7 0
2 years ago
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