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qaws [65]
2 years ago
6

The fuel efficiency for a 2007 passenger car was 31.2 mi/gal. For the same model of car, the fuel efficiency increased to 35.6 m

i/gal in 2012. The gas tank for this car holds 16 gallons of gas. Write and graph a linear function that models the distance that each car can travel for a given amount of gas (up to one tankful). Write the domain and range of each function using interval notation. Write and simplify a function f(g) that represents the difference in the distance that the 2012 car can travel and the distance that the 2007 car can travel on the same amount of gas. Interpret this function using the graphs of the functions from part a. Also, find and interpret f(16). Write the domain and range of the difference function using set notation.

Mathematics
1 answer:
Ierofanga [76]2 years ago
3 0

We have been given that fuel efficiency for a 2007 passenger car was 31.2 mi/gal and the same model of car, the fuel efficiency increased to 35.6 mi/gal in 2012. Also, the gas tank for this car holds 16 gallons of gas.

We need to write a function and graph a linear function that models the distance that each car can travel for a given amount of gas up to one tankful.

Let represent the functions as d_{1}(x) and d_{2}(x)  where d_{1} and d_{2} represent the distances traveled by car in years 2007 and 2012 and x represents the number of gallons. Therefore, we can express the required functions as:

d_{1}(x)=31.2x \text{ and } d_{2}(x) = 35.6x

Domain of both these functions are [0,16] and ranges are [0,499.2] and [0,569.6] respectively for years 2007 and 2012.

The difference function will be:

f(x)=35.6x-31.2x\\
f(x)=4.2x

f(16)=4.2*16=67.2

Domain of this function is [0,16] and range is [0,67.2].

The graphs are shown below.


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WILL GIVE BRAINLIEST AND 39 POINTS
Mashcka [7]

Part 1)
Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 51 km/h. After five hours, the velocity of the car is 59 km/h.

Part 1 a): Write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used

Let

A(3,51) B(5,59)

x------ > represent different times

y------ > represent the velocity of the car

Step 1

Find the slope AB

m=(y2-y1)/(x2-x1)------ > m=(59-51)/(5-3)------ > m=8/2---- > m=4

Step 2

With m=4 and point A(3,51) find the equation of the line

y-y1=m*-(x-x1)------ > y-51=4*(x-3)----- > y=4x-12+51----- > y=4x+39

we know that

The standard form of line equation is Ax + By = C

So

y=4x+39----- > y-4x=39------ > this is the standard form

the answer part 1 a) is

y-4x=39


Part 1 b) How can you graph the equation obtained in Part a) for the first six hours?

To graph the equation obtained in Part a) plot the point A and the point B
and join the points to draw the line


To obtain the velocity for the first six hours, substitute the value of x=6 hour in the equation

for x=6 hour

y-4x=39------ > y-4*6=39------ > y=39+24------ > y=63 km/h


using a graph tool

see the attached figure N 1


Part 2)

g(x)=1+1.5^x

step 1

find the equation of the line of f(x)

let

A(-5,3) B(-3,-1)

m=(-1-3)/(-3+5)----- > m=-4/2---- > m=-2

with m=-2 and point A

y-y1=m*(x-x1)------ > y-3=-2*(x+5)---- > y=-2x-10+3----- > y=-2x-7

so

f(x)=-2x-7

step 2

find the equation of the line of p(x)

let

C(0,2) D(-2,-3)

m=(-3-2)/(-2-0)----- > m=-5/-2---- > m=2.5

with m=2.5 and point C

y-y1=m*(x-x1)------ > y-2=2.5*(x-0)---- > y=2.5x+2

so

p(x)=2.5x+2

Part 2 a) What is the solution to the pair of equations represented by p(x) and f(x)?

We know that

The solution is the intersection of both graphs

Using a graph tool

See the attached figure N 2

The solution is the point (-2,-3)


Part 2 b) Write any two solutions for f(x).

f(x)=-2x-7


for x=0

f(0)=2*0-7---- > f(0)=-7

solution 1 is the point (0,-7)


for x=1

f(1)=2*1-7---- > f(1)=-5

solution 2 is the point (1,-5)


Part 2 c) What is the solution to the equation p(x) = g(x)?

We have

p(x)=2.5x+2

g(x)=1+1.5^x

We know that

The solution is the intersection of both graphs

Using a graph tool

See the attached figure N 3

The solution are the points (0,2) and (7.3,20.2)


Part 3
)

Part A:There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.

Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.

An example of such system of equation is

x < 0

y > 0

The system of equation above represent all the points in the second quadrant of the coordinate system.The area above the x-axis and to the left of the y-axis is shaded.

see the attached figure N 4

Part B:It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.

Substituting D(-4, 2) into the system

we have:

-4 < 0

2 > 0

as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.

Also,

substituting E(-1, 5) into the system we have:

-1 < 0

5 > 0

as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.

Part C:Given that chicken can only be raised in the area defined by y > 3x - 4.

To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.

For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true

For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false

For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false

For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true

For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true

For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true

Therefore,

the farms in which chicken can be raised are the farms at point A, D, E and F.

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