Will give brainliest and 39 points

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 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. (5 points)

part b: how can you graph the equation obtained in part a for the first six hours? (5 points)

question 3 (essay worth 10 points)
(05.02, 5.03 mc)

the graph shows the functions f(x), p(x), and g(x):

graph of function g of x is y is equal to 1 plus the quantity 1.5 raised to the power of x. the straight line f of x joins ordered pairs negative 5, 3 and negative 3, negative 1 and is extended on both sides. the straight line p of x joins the ordered pairs 0, 2 and negative 2, negative 3 and is extended on both sides.
courtesy of texas instruments

part a: what is the solution to the pair of equations represented by p(x) and f(x)? (3 points)

part b: write any two solutions for f(x). (3 points)

part c: what is the solution to the equation p(x) = g(x)? justify your answer. (4 points)

question 4 (essay worth 10 points)
(05.06 hc)

the coordinate plane below represents a town. points a through f are farms in the town.

graph of coordinate plane. point a is at 1, 3. point b is at 3, 1. point c is at 3, negative 3. point d is at negative 4, 2. point e is at negative 1, 5. point f is at negative 3, negative 3.

part a: using the graph above, create a system of inequalities that only contains points d and e in the overlapping shaded regions. explain how the lines will be graphed and shaded on the coordinate grid above. (5 points)

part b: explain how to verify that the points d and e are solutions to the system of inequalities created in part a. (3 points)

part c: chickens can only be raised in the area defined by y > 3x − 4. explain how you can identify farms in which chickens can be raised. (2 points)

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.