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14 days ago

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Which function has no horizontal asymptote? F (x) = StartFraction 2 x minus 1 Over 3 x squared EndFraction f (x) = StartFraction

x minus 1 Over 3 x EndFraction f (x) = StartFraction 2 x squared Over 3 x minus 1 EndFraction f (x) = StartFraction 3 x squared Over x squared minus 1 EndFraction

2 answers:

lawyer [9.2K]14 days ago

8 0

The function has a horizontal asymptote, indicating that it does not continue to infinity because the degree of the numerator exceeds that of the denominator.

tester [8.8K]14 days ago

3 0

Response:

$f(x)=\frac{2x^2}{3x-1}$

Detailed explanation:

The function $f(x)=\frac{2x-1}{3x}$ features a horizontal asymptote which is $y=\frac{2}{3}$

The function $f(x)=\frac{x-1}{3x}$ presents a horizontal asymptote which is $y=\frac{1}{3}$

One function $f(x)=\frac{2x^2}{3x-1}$ lacks a horizontal asymptote because the degree of its numerator surpasses that of the denominator,

The function $f(x)=\frac{3x^2}{x^2-1}$ displays a horizontal asymptote which is $y=3$

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Eric is studying people's typing habits. He surveyed 525 people and asked whether they leave one space or two spaces after a per

lawyer [9240]

Response: (0.8115, 0.8645)

Step-by-step outline:

Define p as the proportion of individuals who leave one space after a sentence.

Provided: Sample size: n= 525

Number of respondents indicating they leave one space: 440

Thus, the sample proportion is: $\hat{p}=\dfrac{440}{525}\approx0.838$

The z-score for a 90% confidence interval is: 1.645

The formula for determining the confidence interval:

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$0.838\pm (1.645)\sqrt{\dfrac{0.838(1-0.838)}{525}}\\\\=0.838\pm (1.645)\sqrt{0.00025858285}\\\\=0.838\pm (1.645)(0.01608)\\\\= 0.838\pm0.0265\\\\=(0.838-0.0265,\ 0.838+0.0265)\\\\=(0.8115,\ 0.8645)$

Consequently, a 90% confidence interval for the proportion of people who leave one space after a period is: (0.8115, 0.8645)

3 0

5 days ago

Five thousand tickets are sold at $1 each for a charity raffle. Tickets are to be drawn at random and monetary prizes awarded a

babunello [8423]

Response:

the expected value of this raffle if you purchase 1 ticket = -0.65

Breakdown of the calculation:

Details:

5,000 tickets are sold at $1 each for a charitable raffle

Winners will be chosen at random with cash prizes as follows: 1 prize of $500, 3 prizes of $300, 5 prizes of $50, and 20 prizes of $5.

Therefore, the value and its respective probability can be calculated as follows:

Value Probability

$500 - $1 = $499 1/5000

$300 - $1 = $299 3/5000

$50 - $1 = $49 5/5000

$5 - $1 = $4 20/5000

-$1 1 - 29/5000 = 4971/5000

The expected value of the raffle when buying 1 ticket is computed as follows:

$E(x) = \sum x * P(x)$

$E(x) = (499 * \dfrac{1}{5000} + 299 *\dfrac{3}{5000} + 49 *\dfrac{5}{5000} + 4 * \dfrac{20}{5000} + (-1 * \dfrac{4971}{5000} ))$

$E(x) = (0.0998 + 0.1794+0.049 + 0.016 + (-0.9942 ))$

$E(x) = (0.3442 -0.9942 )$

$\mathbf{E(x) = -0.65}$

So, the expected value of this raffle when one ticket is purchased = -0.65

7 0

7 days ago

Students have organized

tester [8842]

Answer:

Part A. At least 6 hours

Part B. In less than 2.5 hours, Elijah will fall behind Mercedes

Part C. In over 2.5 hours, Elijah will lead Aubrey

Step-by-step explanation:

D = distance

v = speed

t = time

Formula connecting D, v, and t:

$D=v\cdot t$

Part A.

Steve's speed: $v=3.5\ mph$

Distance: a minimum of 21 miles

Time: unknown, so

$3.5\cdot t\ge 21\\ \\35t\ge 210\ [\text{Multiplied by 10}]\\ \\t\ge \dfrac{210}{35}\\ \\t\ge \dfrac{30}{5}\\ \\t\ge 6$

It would require Steve a minimum of 6 hours to traverse at least 21 miles on Day 1.

Part B.

Mercedes's speed: $v_M=2.4\ mph$

Elijah's speed: $v_E=3.2\ mph$

Elijah's Distance walked: $D_E$ miles

Mercedes's Distance walked: $D_M$ miles

Time: x hours

Mercedes is 2 miles ahead, therefore

$D_E=3.2x\\ \\D_M=2.4x+2$

Elijah will trail behind until

$D_E$

In 2.5 hours, Elijah will close the gap on Mercedes, and in less than 2.5 hours, Elijah will trail behind her.

Part C.

Aubrey's speed: $v_M=3\ mph$

Elijah's speed: $v_E=3.2\ mph$

Elijah's Distance walked: $D_E$ miles

Aubrey's Distance walked: $D_A$ miles

Time: x hours

At the beginning of Day 3, Elijah starts from Mile 42, while Aubrey begins at Mile 42.5.

$D_E=42+3.2x\\ \\D_A=42.5+3x$

Elijah will be ahead of Aubrey when

$D_E>D_A\\ \\42+3.2x> 42.5+3x\\ \\3.2x-3x>42.5-42\\ \\0.2x>0.5\\ \\2x>5\ [\text{Multiplied by 10}]\\ \\x>\dfrac{5}{2}\\ \\x>2.5\ hours$

In 2.5 hours, Elijah will surpass Aubrey, and after more than 2.5 hours, Elijah will outpace Aubrey.

4 0

1 month ago

3. A CD costs £9.50 in London and

PIT_PIT [9121]

Detailed explanation:

Question 4 provides clear information,

thus resolving Question 4:

1 euro equals $1.17.

Consequently, €20.46 is calculated as 20.46 multiplied by 1.17, which equals $23.93.

1 pound equals $1.28.

Thus, £12.60 translates to 12.60 multiplied by 1.28, yielding $16.128.

Therefore, the MP3 in pounds is $7.81 cheaper in dollars.

6 0

18 days ago

Which function has a simplified base of 4RootIndex 3 StartRoot 4 EndRoot? f(x) = 2(RootIndex 3 StartRoot 16 EndRoot) Superscript

tester [8842]

Response:

$f(x)=4\sqrt[3]{16}^{2x}$

Detailed explanation:

You're likely in search of a function with a base that can be simplified to...

$4\sqrt[3]{4}\approx 6.3496$

The functions you seem to be considering appear to be...

$f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x$

It looks like the third option is the one that fits your requirements.

9 0

2 days ago

Read 2 more answers