Solve the inequality.
|3x+7| <_ 4

Using ∣x∣>a⇒x<−a or x>a, we get

∣3x−7∣>4⇒3x−7<−4 or 3x−7>4

⇒3x<3 or 3x>11⇒x<1 or x>

3

11

⇒x∈(−∞,1)∪(

3

11

,∞).

Step-by-step explanation:

it can be written as. -4<_(3x+7)<_4

= -11 <_(3x) <_ (-3)

= (-11/3) <_ X <_ (-1)

Assume that the velocity is a linear function of time, so that
v(t) = a + bt
where
v = velocity (km/h) at time t (hours),
a and b are constants.

When t = 2 hours, v = 18 km/h, therefore
a + 2b = 18         (1)

When t = 4 hours,  v = 4 km/h, therefore
a + 4b = 4          (2)

Subtract (1) from (2) to obtain
a + 4b - (a + 2b) = 4 - 18
2b = -14
  b = -7
From (1), obtain
a = 18 - 2b = 18 - 2*(-7) = 32

Part A.
The equation is
v = 32 - 7t

Part B
To graph the equation for the first 8 hours, create a table as shown below.
t, hours:   0    1    2   3   4   5    6    7    8
v, km/h:  32 25  18  11   4  -3  -10 -17 -24

We can see from the table that the velocity becomes negative between t=4 and t=5. Therefore, we shall change all negative velocities to zero, and assume that the cyclist comes to rest.
TNote that when the velocity is zero,
 32 - 7t = 0
 7t = 32
t = 4.57

The new table becomes
t:    0   1    2   3   4  4.57  5   6   7   8
v: 32 25  18  11   4   0      0   0   0   0
 
The graph is shown below.