Mathematics, 02.08.2019 00:30 nickocasamplonp6mlob

At the smith, inc. company, the number of employees needed to work per year is modeled by the function f(x). f(x)=5001+19e−0.6x which statements are true about the number of employees? a. the company started with 20 employees. b. the maximum number of employees for the company is 500. c. the third year is the first year in which the company will have more than 100 employees. d. the rate of the number of employees per year increases for all values of x.

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The original question gives the fllowing function:
$f(x)=\frac{500}{1+19e^{-0.6x}}$
Let's tackle each option one by one.
For the fisrt statement, the initial condition is found by replacing x with 0.
That is:
$f(0)=\frac{500}{1+19e^{0}}=\frac{500}{20}=25$
Which makes the first statement false.
For the second one we need to take the limit to infinty to see is the functon reaches a maximum:
$\lim_{x\to \infty}f(x)=\frac{500}{1+19\times0}=\frac{500}{1}=500$
Which makes the statemnt correct.
For the thid one we replace x with 2, and thn x with 3 to see if 3 is indeed the firs year that the company will have more than 100 employees:
$f(2)\approx74 \\ f(3)\approx120$
Which makes the statement true.
The fourth statementis checked by takin the second derivative of the function, if the second derivative is negative then the rate is decreasing for all x, if it's possitive the rate is increasing for all x, otherwise the rate might increase or decrase for certain values.

$(f(x))'=500\frac{-0.6.19e^{-0.6x}}{(1+e^{-06x})^2}$
now using the derivative of a rationalfunction formula:
$(\frac{u}{v})'=\frac{u'v-uv'}{v^2}$
We get(after simplifying a lot):
$(f(x))''=5700\frac{e^{-1.2x}(-0.6e^{0.6x}+11.4)}{(1+19e^{-0.6x})^3}$
Which will be negative or high values of x. Thus the fourth statement is false.

Answer from: Quest

B.the point lies directly on the regression

Answer from: Quest

a. we have two

lines: y = 4-x and

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b. 4-x = 8-x^-1

x 4-x 8-x^-1

-3 7

8.33

-2 6 8.5

-1 5 9

0 4

1 3

2 2

7.5

3 1 7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:

x=-4.236

y = 4-x = 4 – (-4.236)

= 8.236

y = 8-x^-1 = .236)^-1

= 8.236

x=0.236

y = 4-x = 4 – (0.236) = 3.764

y = 8-x^-1 = 8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c. to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x and y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines. the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

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