A new hospital tracked the number of births during its first 6 months of operation, as shown in the table.

A 2-column table with 6 rows. The first column is labeled month with entries 1, 2, 3, 4, 5, 6. The second column is labeled number of births with entries 137, 110, 93, 70, 65, 77.
Find a quadratic function that models the data. Round numerical values to the nearest whole number.

f(x) =
x2 +
x +

Use the function to predict the number of births for month 8. There will be about
births.

h df tyjgtyj ttd y ytkgfty ty

step-by-step explanation:

tyty jry j y yj yttj yhgtyj y ghyj ytthj

answer: learning objective(s)

·         describe the creation and use of systems of equations.

·         graph a system of linear equations on the coordinate plane and identify its solution.

introduction

sometimes graphing a single linear equation is all it takes to solve a mathematical problem. other times, one line just doesn’t do it, and a second equation is needed to model the situation. this is often the case when a problem involves two variables. solving these kinds of problems requires working with a system of equations, which is a set of two or more equations containing the same unknowns.

let’s take a look at systems of equations, and see what the graphs of individual equations within a system reveal about the mathematical relationship of the variables.

systems of equations

a system of equations contains two or more linear equations that share two or more unknowns. to find a solution for a system of equations, we must find a value (or range of values) that is true for all equations in the system.

the graphs of equations within a system can tell us how many solutions exist for that system. look at the images below. each show two lines that make up a system of equations (in the graph on the right the two lines are superimposed and look like a single line). how many points in common does each system of lines reveal?

one solution

no solutions

infinite solutions

if the graphs of the equations intersect, then there is one solution that is true for both equations.

if the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.

if the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

remember, the graph of a line represents every point that is a possible solution for the equation of that line. so when the graphs of two equations cross, the point of intersection lies on both lines, meaning that it is a possible solution for both equations. when the graphs of two equations never touch, there are no shared points and there are no possible solutions for the system. when the graphs of two equations lie on top of one another, they share all their points and every one is a possible solution.

graphing as a solution method

graphing equations in order to identify a specific point of intersection is usually not a precise way to solve systems because it is often difficult to see exactly where two lines intersect (unless you are using a computer-based graphing program that allows you to zoom in on a point). however, the graph of a system of equations can still give a good idea of what type of solution, if any, exists. let’s graph a system, and see how it works.

graph the system y = 3x and x + 2y = 4. how many solutions does this system have?

a graph of the two lines y = 3x and x + 2y = 4 shows us that the lines intersect, meaning that there is a single (x, y) value that satisfies both equations. looking at the graph does not tell us exactly where that point is, but we don’t need to know that information here, because the question only asks for the number of solutions.

so a system made of two intersecting lines has one solution. now let’s look at a different situation:

how many solutions exist for the system y -0.5x = 7 and 2y = x − 3?

plotting both equations, it looks like there is no solution—the lines are parallel. to check this finding, we can compare the slopes of the equations. to make that easy, we’ll rearrange the equations into slope-intercept form, or y = mx + b.   that gives us the equations y = 0.5x + 7 and y = 0.5x − 1.5.   yes, the slope of both lines is 0.5, which means the lines really are parallel. they never intersect, so there is no point that lies on both lines, and no solution to the system.

step-by-step explanation:

answer: procent 54 wynosi 33,33%

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).