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2 years ago

Next you need to find the critical points. What are the critical points, and how are they found?

Mathematics

2 answers:

m_a_m_a [10]2 years ago

8 0

For x-25/5(10+x) less than or equal to 0 .

The critical points are found by setting the numerator and denominator to equal 0.

The critical points are at -10 and 25.

Sophie [7]2 years ago

6 0

Answer:

The critical points are found by setting the numerator and denominator equal to 0.

The critical points are at –10 and 25

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In a lecture demonstration, an object is suspended from a spring scale which reads 8N when the object is in air. The object is t

katovenus [111]

Answer:

a) density of the object is 3995.01, b) the weight scale reads 22N c) the sum individually will be the same with when added together.

Step-by-step explanation:

The weight of the object in air is 8N,

and weight = Mass * acceleration due to gravity = m * 9.81

8/9.81 = 0.815,

upthrust( force acting on the body from the liquid impeding the immersion) on the body when fully submerged = weight in air - weight in water = 8N - 6N =2N

Upthrust = weight of water displaced = 2N = mass * acceleration

2/9.81 = 0.204kg

density of water(1000kg/m^3) = mass of water / volume of water

volume of water displaced = 0.204/1000 = 0.000204m^3 (204cm^3)

volume of water displaced = volume of the solid

density of solid = mass/ volume = 0.815/0.000204 = 3995.01kg/m^3

b) when fully submerge in water the the scale experience according to newton third law of motion ( equal and opposite reaction of forces) additional 2N push so that total weight with the fully submerge solid = 20N + upthrust = 20N + 2N =22N

c) the of two scale reading is before (8N + 20N = 28N) and after (6N + 22N = 28) since there is no loss of matter; the demonstration was in equilibrium.

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2 years ago

A publishing company prints newspapers and magazines. Let NNN represent the number of newspapers and MMM represent the number of

adelina 88 [10]

Answer:

At most 800 magazines the company can print daily with the remaining number of ink cartridges.

Step-by-step explanation:

We are given the following in the question:

$0.5N+2.5M \leq 6000$

The above inequality gives the relation for daily supply of ink cartridges where N is the number of newspaper and M is the number of magazines.

Number of newspaper to be printed daily,N = 8000

We have to find the number of magazines at most can the company print daily with the remaining number of ink cartridges.

Puting the value in the given inequality,

$0.5(8000)+2.5M \leq 6000\\\Rightarrow 4000+2.5M \leq 6000\\\Rightarrow 2.5M \leq 2000\\\\\Rightarrow M \leq \dfrac{2000}{2.5}\\\\\Rightarrow M \leq 800$

Thus, at most 800 magazines the company can print daily with the remaining number of ink cartridges.

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2 years ago

How far should Galileo walk up the inclined plane?

sergejj [24]

Answer:

Step-by-step explanation:

150/x=sin 15

x=150/sin 15 ≈579.56 m

4 0

2 years ago

Which graph represents the piecewise-defined function?

chubhunter [2.5K]

Answer: The correct option is 1.

Explanation:

The given piecewise function is,

$y=\begin{cases}-\frac{4}{5}x-3 & \text{ if } x$

It means if x<0, then

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

If $x\geq2$ , then

$f(x)=3x-10$

Since the f(x) is defined for x<0 and $x\geq2$ , therefore the function f(x) is not defined for $0\leq x$ .

From the graph 2, 3 and 4 we can easily noticed that for each value of x there exist a unique value of y, therefore the function is defined for all values of x, which is not true according to the given piecewise function.

Only in figure the value of y not exist when x lies between 0 to 2, including 0. It means the function is not defined for $0\leq x$ , hence the first option is correct.

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2 years ago

Consider sampling heights from the population of all female college soccer players in the United States. Assume the mean height

BARSIC [14]

Answer:

B) .35

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .