Select the sets that are not functions.

5. Which function would have the point (0,0) on its graph?

stepladder [879]

Answer:

B. Output = Input x 5

Step-by-step explanation:

Given:

Point (0,0)

To find which of the functions given will have the point (0,0) on its graph.

Solution:

In order to check whether a point $(x,y)$ lies on the graph of the function, we plugin the $x$ as the input in the function and $y$ as the output of the function and see if it satisfies the equation.

Plugging in point (0,0) in the functions given:

A. Output = Input + 3

Plugging in Output=0 and Input=0

$0=0+3$

$0=3$

The above statement can never be true and hence, the function graph does not have the point (0,0).

B. Output = Input x 5

Plugging in Output=0 and Input=0

$0=0\times 5$

$0=0$

The above statement is always true and hence, the function graph does have the point (0,0).

C. Output = Input - 2

Plugging in Output=0 and Input=0

$0=0-2$

$0=-2$

The above statement can never be true and hence, the function graph does not have the point (0,0).

D. Output = Input + 3

Plugging in Output=0 and Input=0

$0=0+3$

$0=3$

The above statement can never be true and hence, the function graph does not have the point (0,0).

4 0

3 years ago

In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te^(−2t)−(8t+12) with

Zarrin [17]

First check the characteristic solution:

y'' + 4y' + 4y = 0

has characteristic equation

r ² + 4r + 4 = (r + 2)² = 0

with a double root at r = -2, so the characteristic solution is

$y_c = C_1e^{-2t} + C_2te^{-2t}$

For the particular solution corresponding to $12te^{-2t}$ , we might first try the ansatz

$y_p = (At+B)e^{-2t}$

but $e^{-2t}$ and $te^{-2t}$ are already accounted for in the characteristic solution. So we instead use

$y_p = (At^3+Bt^2)e^{-2t}$

which has derivatives

${y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}$

${y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}$

Substituting these into the left side of the ODE gives

$(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}$

so that 6A = 12 and 2B = 0, or A = 2 and B = 0.

For the second solution corresponding to $-8t-12$ , we use

$y_p = Ct + D$

with derivative

${y_p}' = C$

${y_p}'' = 0$

Substituting these gives

$4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12$

so that 4C = -8 and 4C + 4D = -12, or C = -2 and D = -1.

Then the general solution to the ODE is

$y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1$

Given the initial conditions y (0) = -2 and y' (0) = 1, we have

$-2 = C_1 - 1 \implies C_1 = -1$

$1 = -2C_1 + C_2 - 2 \implies C_2 = 1$

and so the particular solution satisfying these conditions is

$y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1$

or

$\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}$

7 0

2 years ago

What is the slope when it passes through 6,3 and 2,8

pantera1 [17]

Answer: m=-5/4

Step-by-step explanation:

4 0

3 years ago

Identify whether the series summation of 12 open parentheses 3 over 5 close parentheses to the i minus 1 power from 1 to infinit

Paraphin [41]

Answer:

This is a convergent geometric series. The sum is 30.

Step-by-step explanation:

In your series, the ratio of terms is 3/5, a value whose magnitude is less than 1. That means the series is convergent. The sum is given by the formula ...

S = a1/(1 -r)

where a1 is the first term (12) and r is the common ratio (3/5). This gives ...

S = 12/(1 -3/5) = 12·5/2 = 30

The series is convergent and has a sum of 30.

7 0

3 years ago

Annual salary plus bonus data for chief executive officers are presented in the BusinessWeek Annual Pay Survey. A preliminary sa

Anuta_ua [19.1K]

Answer:

175

Step-by-step explanation:

The sample size n in Simple Random Sampling is given by

$\bf n=\frac{z^2s^2}{e^2}$

where z = 1.96 is the z-score associated with a 95% confidence level ans s the standard deviation.

Replacing values

$\bf n=\frac{(1.96)^2(675)^2}{(100)^2}=175$

So, there should be 175 chief executive officers in a sample if we want to estimate the population mean annual salary plus bonus with a margin of error of $100,000

7 0

3 years ago