All categories

1 year ago

The range of which function is (2, infinity)? y = 2x y = 2(5x) y = 5x +2 y = 5x + 2

Mathematics

2 answers:

Arturiano [62]1 year ago

5 0

Answer:

y = 5∧× + 2

Step-by-step explanation:

Sveta_85 [38]1 year ago

3 0

Answer:

I think your functions are $y=2^{x}$ , $y=2*5^{x}$ and $y=2+5^{x}$

If yes then then the third function which is $y=2+5^{x}$ .

Step-by-step explanation:

The function $c^{x}$ where c is a constant has

Domain : $c\geq 0$

Range : ( 0 , ∞ )

The above range is irrespective of the value of c.

I have attached the graph of each of the function, you can look at it for visualization.

$y=2^{x}$ ⇒ This function is same as $c^{x}$ so its range is ( 0 , ∞ ).

$y=2*5^{x}$ ⇒ If we double each value of the function $y=5^{x}$ , which has range ( 0 , ∞ ), but still the value of extremes won't change as 0*2=0 and ∞*2=∞. Therefore the range remains as ( 0 , ∞ ).

$y=2+5^{x}$ ⇒ If we add 2 to each value of the function $y=5^{x}$ , which has range ( 0 , ∞ ), the lower limit will change as 0+2=2 but the upper limit will be same as ∞. Therefore the range will become as ( 2 , ∞ ).

You might be interested in

Suppose that a manager is interested in estimating the average amount of money customers spend in her store. After sampling 36 t

musickatia [10]

Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Step-by-step explanation:

Assuming this complete question: Which statement gives a valid interpretation of the interval?

The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.

There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.

There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The 90% confidence interval for this case would be (38.01, 44.29) and is given.

The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .

And the best option would be:

The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29

8 0

2 years ago

Which sentences are punctuated correctly? Select 2 options.

baherus [9]

Answer:

Step-by-step explanation:

Andrew's favorite vegetables are cucumbers, green beans, and red peppers.

Samantha enjoyed many types of music; jazz, classical, hip hop, country, and heavy metal,

Correct me if I'm wrong

7 0

2 years ago

Read 2 more answers

Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i

jenyasd209 [6]

Answer:

If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

$\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}$

Observe that

$\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}$

and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

$\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}$

3 0

1 year ago

A jet flying at 123 m/s banks to make a horizontal circular turn. The radius of the turn is 3810 m, and the mass of the jet is 2

aev [14]

Answer:

The magnitude of the necessary lifting force is 7.95N

Step-by-step explanation:

Force = mv^2/r

mass (m) = 2.00105kg, velocity (v) = 123m/s, radius (r) = 3810m

Force = 2.00105×123^2/3810 = 30273.89/3810 = 7.95N

7 0

1 year ago

Chen is bringing fruit and veggies to serve at an afternoon meeting. He spends a total of $28.70 on 5 pints of cut veggies and 7

deff fn [24]

The answer would be $1.75 (Rounded off to the nearest cent).

Substitute the given cost of each pint of fruit to the equation. Transpose and simplify, to get the price of a pint of veggies.
Given: 5v + 7f = 28.70; f = $2.85
Solution: 5v + 7(2.85) = 28.70
5v + 19.95 = 28.70
5v = 8.75
v = 1.75

6 0

2 years ago

Read 2 more answers