I need help asap!! thanks, 15 pts

the 2nd and 3rd

Step-by-step explanation:

3x-4>5                                                      -8x-19<21

   +4   +4                                                         +19   +19

3x>9                                                              -8x<40

---    ---                                                              

3      3                                                               -8     -8

x<3                                                                    x>-5

V = -5600t + 148,950

Step-by-step explanation:

In this question, we are expected to give an equation or relation that relates the value of a bulldozer V to the time it took for it to have depreciated to that particular value.

This question is seeking to establish a linear model to relate the depreciated value V to the time taken t.

Now, since it is a linear model, let’s picture a plot. A graphical plot that has time t on the x axis and Value V on the y axis.

At year 0, when the bulldozer was newly bought, it has a value of $148,950.

Writing this in form of coordinates, we can say that the point here on the Cartesian plane would be (0, 148,950)

Now, after 24 years, depreciated value is $14,550. This means that the point we could have here is (24, 14,550)

Now, we can calculate the slope of the line to be modeled using these two points;

Gradient of linear line= (14,550-148,950)/(24 - 0) = -5,600

Mathematically, the equation of a linear model is y = mx + c

Where m is the gradient of the line and c is the y intercept of the line. Here, we can identify the gradient as -5,600 , y-intercept as value of bulldozer at when t is 0 years, y is our value V and x is our time t.

Putting the equation together, we have;

V = -5600t + 148,950

We should understand that our slope is negative because, the value of the bulldozer is expected to decrease as the years toll by