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BARSIC [14]
2 years ago
6

What are the domain, range, and asymptote of h(x) = 6x – 4? domain: {x | x is a real number}; range: {y | y > 4}; asymptote:

y = 4 domain: {x | x is a real number}; range: {y | y > –4}; asymptote: y = –4 domain: {x | x > –4}; range: {y | y is a real number}; asymptote: y = 4 domain: {x | x > 4}; range: {y | y is a real number}; asymptote: y = –4
Mathematics
2 answers:
Sladkaya [172]2 years ago
6 0

Answer:

the answer is b

Step-by-step explanation:

Mkey [24]2 years ago
4 0

Answer:

Step-by-step explanation:

The domain of a function is the set for which the function is defined. Our function is the function h(x) = 6x-4. This function is defined regardless of the value of x, so it is defined for every real value of x. That is, it's domain is the set {x|x is a real number}.

The range of the function is the set of all possible values that the function might take, that is {y|y=6x-4}. Recall that every real number y could be written of the form y=6x-4 for a particular x. So the range of the function is the set {y|y is a real number}.

Note that as x gets bigger, the value of 6x-4 gets also bigger, then it doesn't approach any particular number. Note also that as x approaches - infinity, the value of 6x-4 approaches also - infinity. In this case, we don't have any horizontal asymptote. Since the function is defined for every real number, it doesn't have any vertical asymptote. Since h is a linear function, it cannot have any oblique asymptote, then h doesn't have any asymptote.

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The area of a rectangle 3 yd by 10 yd is
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2 years ago
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A major traffic problem in the Greater Cincinnati area involves traffic attempting to cross the Ohio River from Cincinnati to Ke
myrzilka [38]

Answer:

a) 0.36

b) 0.3

c) Yes

Step-by-step explanation:

Given:

Probability of no traffic delay in one period, given no traffic delay in the preceding period = P(No_Delay) = 0.9

Probability of finding a traffic delay in one period, given a delay in the preceding period = P(Delay) = 0.6

Period considered = 30 minutes

a)

Let A be the probability that for the next 60 minutes (two time periods) the system will be in the delay state:

As the Probability of finding a traffic delay in one period, given a delay in the preceding period is 0.6 and one period is considered as 30 minutes.

So probability that for the next two time periods i.e. 30*2 = 60 minutes, the system in Delay is

P(A) = P(Delay) * P(Delay) =  0.6 * 0.6 = 0.36

b)

Let B be the probability that in the long run the traffic will not be in the delay state.

This statement means that the traffic will not be in Delay state but be in No_Delay state in long run.

Let C be the probability of one period in Delay state given that preceding period in No-delay state :

P(C) =  1 - P(No_Delay)

        = 1 - 0.9

P(C) = 0.1

Now using P(C) and P(Delay) we can compute P(B) as:

P(B) = 1 - (P(Delay) + P(C))

      = 1 - ( 0.6 + 0.10 )

      = 1 - 0.7

P(B) = 0.3

c)

Yes this assumption should be questioned for this traffic problem because it implies that  traffic will be in Delay state for the 30 minutes and just after 30 minutes, it will be in No_Delay state. However, traffic does not work like this in general and it makes this scenario unrealistic. Markov process model can be improved if probabilities are modeled as a function of time instead of being presented as constant (for 30 mins).

8 0
2 years ago
Gianna is going to throw a ball from the top floor of her middle school. When she throws the ball from 32feet above the ground,
Jlenok [28]

Answer:

There are two times for the ball to reach a height of 64 feet:

1 second after thrown ⇒ the ball moves upward

2 seconds after thrown ⇒ the ball moves downward

Step-by-step explanation:

* Lets explain the function to solve the problem

- h(t) models the height of the ball above the ground as a function

 of the time t

- h(t) = -16t² + 48t + 32

- Where h(t) is the height of the ball from the ground after t seconds

- The ball is thrown upward with initial velocity 48 feet/second

- The ball is thrown from height 32 feet above the ground

- The acceleration of the gravity is -32 feet/sec²

- To find the time when the height of the ball is above the ground

  by 64 feet substitute h by 64

∵ h(t) = -16t² + 48t + 32

∵ h = 64

∴ 64 = -16t² + 48t + 32 ⇒ subtract 64 from both sides

∴ 0 = -16t² + 48t - 32 ⇒ multiply the both sides by -1

∴ 16t² - 48t + 32 = 0 ⇒ divide both sides by 16 because all terms have

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∴ t² - 3t + 2 = 0 ⇒ factorize it

∴ (t - 2)(t - 1) = 0

- Equate each bracket by zero to find t

∴ t - 2 = 0 ⇒ add 2 to both sides

∴ t = 2

- OR

∴ t - 1 = 0 ⇒ add 1 to both sides

∴ t = 1

- That means the ball will be at height 64 feet after 1 second when it

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 moves down

* There are two times for the ball to reach a height of 64 feet

  1 second after thrown ⇒ the ball moves upward

  2 seconds after thrown ⇒ the ball moves downward

3 0
2 years ago
gavin is making cranberry lemonade. he wants to make a total of 6 cups. he needs to use 3 times as much lemonade as cranberry ju
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Answer:

3l= c

Step-by-step explanation:

as use 3 cup of cranberry juice, lemonade will be used 1 cup.

4 0
2 years ago
The "normal" high temperature for a June day in Albemarle is $82^\circ$ (degrees Fahrenheit). Last year, the first ten days of J
crimeas [40]

Answer:

6° F

Step-by-step explanation:

To get average of 10 days highs, we need to add all and divide by 10.

Average = \frac{84+88+91+92+96+94+89+78+87+81}{10}\\=\frac{880}{10}\\=88

And, the normal high temp is given as 82.

<em>How much higher is 88 than 82??</em>

<em />

It is 88 - 82 = 6°F above

8 0
2 years ago
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