For the function y=ln(x-1)+2 which of the following statements is truea. the domain is all real numbers and the range is [2, infinity)
b. the domain is (-1, infintity} and the range is all real numbers
c. the domain is (1, infinity) and the range is [2, infinity)
d. the domain is (1, infinity) and the range is all real numbers

Answers

Answer 1

Answer:

For the function $y = ln(x-1) + 2$ , statement d is true. The domain is (1, ∞) and the range is all real numbers.

A function is an expression or a rule establishing a relationship between two sets or two variables, where one is independent and another is dependent.

The set of values you input into the data as the independent variable is called the domain of the function.

The set of possible outputs of the function, the dependent variable, is called the codomain of the function.

The set of elements part of the dependent variable that actually comes out of the function as output is called the range of the function.

Given function, $y = ln(x-1) + 2$

The domain of the function is what you can put into x.

for ln(x-1) to be defined, x-1 > 0 implies that x > 1

Thus the domain of function becomes (1, ∞).

The range of the function is what you get as y.

if 1 < x < 2, 0 < x-1 < 1, ln(x-1) < 0, thus $y = ln(x-1) + 2$ will have a value y < 2, maybe even negative.

if x = 2, x-1 = 1, ln(x-1) = ln(1) = 0, making y = 2

if x > 2, x-1 > 3, ln(x-1) > 0, making y >2.

Thus the range of function becomes (-∞, ∞).

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Answer 2

Answer:

Answer:

d. the domain is (1, infinity) and the range is all real numbers

Step-by-step explanation:

The domain of a logarithmic function

$f(x)=\ln(x)$

is the set of all positive numbers

$D:\ x>0\Rightarrow x\in\mathbb{R}^+$

The range of a logarithmic function

$f(x)=\ln(x)$

is the set of all real numbers

$R:\ y\in\mathbb{R}$

We have:

$y=\ln(x-1)+2$

DOMAIN

$x-1>0$ add 1 to both sides

$x-1+1>0+1\n\nx>1$

$D:x>0\Rightarrow x\in(1,\ \infty)$

RANGE

$f(x)=\ln(x)\to f(x)+2=\ln(x)+2$

The graph shifted 2 units up. The range no change.

$R:\ y\in\mathbb{R}$

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Suppose a movie starts at 5:00 p.m. and Lindsay, a customer who is always late, arrives at the movie theater at a random time between 5:10 p.m. and 5:45 p.m. Lindsay's late arrival time, in minutes, represented by ???? , models a uniform distribution between 10 and 45 min. Determine the height of the uniform density curve. Provide your answer with precision to three decimal places.

Answers

Answer: The height of uniform density curve is 0.028.

Step-by-step explanation:

Since we have given that

Uniform distribution between 10 and 45 minutes.

Here,

a = 10 minutes

b = 45 minutes

We need to find the height of the uniform density curve.

So, $f(X=x)=(1)/(b-a)=(1)/(45-10)=(1)/(35)=0.028$

So, the height of uniform density curve is 0.028.

-x + 3y = 3

x - 3y = 3

Does this system have a solution?

Answers

Answer:

No solution

Step-by-step explanation:

Slope-Intercept Form: y = mx + b

Step 1: Write out systems of equations

-x + 3y = 3

x - 3y = 3

Step 2: Rewrite equations into slope-intercept form

3y = 3 + x

y = 1 + x/3

-3y = 3 - x

y = -1 + x/3

Step 3: Rewrite systems of equations

y = x/3 + 1

y = x/3 - 1

Since we have the same slope for both equations but different y-intercepts, we know that both lines are parallel. If that is the case, they will never touch or intersect each other. Therefore, we have no solution.

12 times 10 to the 0 power equals 12 times one equals?

Answers

12×10 to the power of zero is 1.2×10 to the power of one and 12 times one is 12

A small jar of peanut butter sells for 0.08 per ounce. A large jar of peanut butter sells for $1.20 per pound. Which is the better buy and by how much (in cents per pound)?

Answers

Answer:

a small jar of penuts buteer sells for 0.08 per ounce

A large jar of penut buttter sells for $1.20 per pound

the answer is :

hope it will help you

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. y 2 ( y 2 − 4 ) = x 2 ( x 2 − 5 ) , ( 0 , − 2 ) (devil's curve) y2(y2-4)=x2(x2-5), (0,-2) (devil's curve)

Answers

Answer:

y = -2

Step-by-step explanation:

To find the equation of the tangent we apply implicit differentiation, and then we take apart dy/dx

The equation is

$y^2(y^2-4)=x^2(x^2-5)$

implicit differentiation give us

$(d)/(dx)[y^2(y^2-4)=x^2(x^2-5)]\n\n2y(dy)/(dx)(y^2-4)+y^2(2y(dy)/(dx))=2x(x^2-5)+x^2(2x)\n\n4y^3(dy)/(dx)-8y(dy)/(dx)=2x^3-10x+2x^3\n\n(dy)/(dx)=(4x^3-10x)/(4y^3-8y)$

But we know that

$m=(dy)/(dx)\ny=mx+b$

Hence, for the point (0,-2) and by replacing for dy/dx

$m=(dy)/(dx)_((0,-2))=(4(0)+10(0))/(4(-2)^3-8(-2))=0$

Hence m=0, that is, the tangent line to the point is a horizontal line that cross the y axis for y=-2. The equation is:

y=(0)x+b = -2

HOPE THIS HELPS!!

In order to find the equation of the tangent line to the curve y²(y² - 4) = x²(x² - 5) at the point (0, -2), we will use the method of implicit differentiation. Here are the steps:

Step 1: Differentiate Each Side of the Given Equation with Respect to x

Applying the chain rule to differentiate y²(y² - 4) with respect to x gives:
2y*y'(y² - 4) + y²*2y*y' = d/dx [y²(y² - 4)]
The chain rule is also applied to differentiate x²(x² - 5) with respect to x, yielding:
2x(x² - 5) + x²*2x = d/dx [x²(x² - 5)]

Step 2: Equate the Two Expressions Found from Step 1 and Solve for y'

2y*y'(y² - 4) + y²*2y*y' = 2x(x² - 5) + x²*2x

This equation can be solved by isolating y' (the derivative of y with respect to x), which represents the slope of the tangent line.

Step 3: Use the Given Point (0, -2) to Find the Slope of the Tangent Line

Substitute x = 0 and y = -2 into the equation found in Step 2 to get the specific value for the slope at the given point.

Step 4: Use the Point-Slope Form of the Line to Write the Equation of the Tangent Line

The point-slope form of the line y - y₁ = m(x - x₁) can be used to write the equation of the tangent line. We substitute for x₁ and y₁ with the coordinates of the given point (0, -2), and m with the slope found from Step 3.

The resulting equation represents the tangent line to the curve at the given point (0, -2). Please note that the full calculation may result in a complex slope due to the nature of the given curve equation. Nonetheless, this process illustrates the application of implicit differentiation and the point-slope form of a line in finding the equation of a tangent line to a curve.

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