Maths-
General
Easy

Question

A road sign shows a Vehicle's speed as the vehicle passes.

a. The sign blinks for vehicles travelling within 5 mi divided by straight h of the speed limit. Write and solve an absolute value inequality to find the minimum and maximum speeds of an oncoming vehicle that will cause the sign to blink.
b. Another sign blinks when it detects a vehicle travelling within 2 mi divided by straight h of a 35 mi divided by straight h  speed limit. Write and solve an absolute value inequality to represent the speeds of the vehicles that cause the sign to blink.

hintHint:

|x| is known as the absolute value of x. It is the non-negative value of x irrespective of its sign. The value of absolute value of x is given by
vertical line x vertical line equals open curly brackets table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell negative x comma x less than 0 end cell row cell x comma x greater or equal than 0 end cell end table close
We will construct an equation for both the given cases which models the situation with the help of the concept of absolute value.

The correct answer is: Hence, we get two values of x satisfying the given equation, x = 33,37


    Step by step solution:
    (a) Given,
    Let the speed of the oncoming vehicle be x mi/h
    Speed limit of the vehicles = 30 mi/h
    Error allowed in the speed limit = plus-or-minus  5 mi/h
    We form an equation in terms of speed of the vehicle.
    The two conditions we have are

     x less or equal than 30 plus 5
    And

     x greater or equal than 30 minus 5
    We can rewrite the above equations as

    x minus 30 less or equal than 5
    And

    x minus 30 greater or equal than negative 5

    not stretchy rightwards double arrow negative left parenthesis x minus 30 right parenthesis less or equal than 5
    We combine the above two cases to get the following equation.
    Thus, the equation representing the given situation is

    vertical line x minus 30 vertical line less or equal than 5
    We solve the above equation to get two values of x, which will be the minimum and maximum values  of x.
    Using the definition of absolute value,

    vertical line x vertical line equals open curly brackets table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell negative x comma x less than 0 end cell row cell x comma x greater or equal than 0 end cell end table close
    We get two possibilities,

    Forx minus 30 less than 0

    vertical line x minus 30 vertical line equals negative left parenthesis x minus 30 right parenthesis equals 5
    Simplifying, we get

     negative x plus 30 equals 5
    Subtracting 30 both sides, we have

     negative x equals 5 minus 30 equals negative 25
    Dividing throughout by -1, we get

     x = 25

    For x minus 30 greater or equal than 0 ,

    vertical line x minus 30 vertical line equals x minus 30 equals 5
    Adding 30 both sides, we get

     x equals 30 plus 5
    Thus, we get

    x = 35
    Hence, we get two values of x satisfying the given equation,

    x = 35,25
    Thus,
    Maximum speed of the oncoming vehicle which makes the sign to blink = 35 mi/h
    Minimum speed of the oncoming vehicle which makes the sign to blink = 25 mi/h
    b) Given,
    Let the speed of the oncoming vehicle be x mi/h
    Speed limit of the vehicles = 35 mi/h
    Error allowed in the speed limit =   2 mi/h
    We form an equation in terms of speed of the vehicle.
    The two conditions we have are

    x less or equal than 35 plus 2
    And

    x greater or equal than 35 minus 2
    We can rewrite the above equations as

    x minus 35 less or equal than 2
    And

    x minus 35 greater or equal than negative 2

    not stretchy rightwards double arrow negative left parenthesis x minus 35 right parenthesis less or equal than 2
    We combine the above two cases to get the following equation.
    Thus, the equation representing the given situation is

     vertical line x minus 35 vertical line less or equal than 2
    We solve the above equation to get two values of x, which will be the minimum and maximum values  of x.
    Using the definition of absolute value,

    vertical line x vertical line equals open curly brackets table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell negative x comma x less than 0 end cell row cell x comma x greater or equal than 0 end cell end table close
    We get two possibilities,

    For ,x minus 35 less than 0,

    vertical line x minus 35 vertical line equals negative left parenthesis x minus 35 right parenthesis equals 2
    Simplifying, we get

    negative x plus 35 equals 2
    Subtracting 35 both sides, we have

     negative x equals 2 minus 35 equals negative 33
    Dividing throughout by -1, we get

     x equals 33

    For x minus 35 greater or equal than 0 ,

    vertical line x minus 35 vertical line equals x minus 35 equals 2
    Adding 30 both sides, we get

    x equals 35 plus 2
    Thus, we get

    x = 37
    Hence, we get two values of x satisfying the given equation,

    x = 33,37

    |x|, which is pronounced "Mod x" or "Modulus of x," stands in for the absolute value of the variable x. The measure is the meaning of the Latin term "modulus." Common names for absolute value include numerical value and magnitude. The absolute value does not include the sign of the numeric value; it solely represents the numeric value. Any vector quantity's modulus is its absolute value and is always assumed to be positive.
    Furthermore, absolute values express all quantities, including time, price, volume, and distance. Take the absolute value as an example: |+5| = |-5| = 5. The absolute value has no assigned sign. The formula to calculate a number's absolute value is |x| = x if it is greater than zero, |x| = -x if it is less than zero, and |x| = 0 if it is equal to zero.

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