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2 years ago

What is the range of the exponential function shown below

Mathematics

2 answers:

kvv77 [185]2 years ago

6 0

Answer:

The correct option is C.

Step-by-step explanation:

Range is the set of output or the values of the function.

The given function is

$f(x)=9\cdot 2^x$

If a exponential function is defined as g(x)=a^x, where a>0, then the value of g(x) is always greater than 0, g(x)>0.

It means,

$2^x>0$

Multiply both sides by 9.

$9\cdot 2^x>9\cdot 0$

$9\cdot 2^x>0$

$f(x)>0$

The value of the function f(x) is always greater than 0, therefore the range of the function is

Range = { y | y>0}

Hence the correct option is C.

Vsevolod [243]2 years ago

3 0

Answer : y>0

f(x) = 9*2^x

f(x) is an exponential function

$f(x) = 9*2^x$

When we plug in positive value for x , the value of y is positive

When we plug in negative value for x , the value y is also positive

So for any value of x, the y value is positive always.

Range is the set of y values for which the function is defined

y values are positive , so range is y >0

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Which statements about the hyperbola are true? Check all that apply.

mariarad [96]

Answer:

True options: 1, 2 and 5

Step-by-step explanation:

From the given diagram, you can see that the center of the hyperbola is placed at the origin, so first option is true (see attached diagram for definition of center, vertices, foci, i.e.)

There are two vertices of the hyperbola, they are placed at (-6,0) and (6,0), so second option is true.

The transverse axis is the segment connecting vertices, this segment is horizontal, so option 3 is false.

The foci are not placed within the rectangular reference box, so this option is false.

The directrices are vertical lines with equations $x=\pm \dfrac{a}{e}$ , so this option is true.

8 0

2 years ago

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In Drew's town the average price for a tutor is $15 per hour, and the standard deviation is $5.25 per hour. In Drew's town, what

alukav5142 [94]

Answer:

0.659 is the probability that a tutor charges between $10 and $20 per hour.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $15 per hour

Standard Deviation, σ = $5.25 per hour

We are given that the distribution of tutoring prices is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P( tutor charges between $10 and $20 per hour)

$P(10 \leq x \leq 20)\\\\ = P(\displaystyle\frac{10 - 15}{5.25} \leq z \leq \displaystyle\frac{20-15}{5.25})\\\\ = P(-0.9523 \leq z \leq 0.9523)\\\\= P(z \leq 0.9523) - P(z < -−0.9523)\\= 0.8295 - 0.1705= 0.659$

0.659 is the probability that a tutor charges between $10 and $20 per hour.

5 0

2 years ago

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

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

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2 years ago

To buy a computer, Raquel borrowed $3,000 at 9%

kumpel [21]

Answer:

$4080

Step-by-step explanation:

We have the amount she will pay back, but first, we need to find the Interest accrued.

Simple Interest is given as:

$I = \frac{P * R * T }{100}$

where P = principal

R = rate

T = time taken (in years)

Therefore, the interest on $3,000 at 9% simple interest for 4 years is:

$I = \frac{3000 * 9 * 4}{100}$

I = $1080

Therefore, the amount she will pay back is:

$3000 + $1080 = $4080

6 0

1 year ago

The means and mean absolute deviations of the amount of rain that fell each day in a local city, last week and this week, are sh

Dennis_Churaev [7]

The difference in the mean rainfall is given by 19.7 - 19.3 = 0.5

The difference in the <span>mean rainfall is approximately 0.5/4.6 = 0.087 = 8.7 percent of the mean absolute deviation for last week and is 0.5/5.2 = 0.077 = 7.7 percent of the mean absolute deviation for this week.

Therefore, the difference in the </span><span>mean rainfall is approximately what percent of the mean absolute deviation 8% of the mean absolute deviation of the data sets.</span>

4 0

2 years ago

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