Answer:
59.0 miles per hour
Step-by-step explanation:
When you have an equation with an absolute value, you must consider two options: one with a positive result and another one with a negative result.

In this case you need the maximum safe driving speed which is 59.2 miles per hour
Answer:

Step-by-step explanation:
Let the number be x
Given

Required
Find x

Collect Like Terms

Take LCM


Multiply both sides by 6


Divide both sides by -3


Hence, the number is 
Answer:
A.-
D.
E.
Step-by-step explanation:
Like terms must have the same variable, in this case x, and the same exponent, in this case 2. Since the original term is
, the like terms will be those that contain
, regardless of whether their coefficient or sign is different.
Analyzing the options:
A.-
We have the same variable and the same exponent
, so it is a like term.
B. 
You have the same variable x but not the same exponent. So it's not a like term of 
C.
Same variable
but as in the previous case, the exponent is different, it is a 3 and it should be a 2, so it is not a similar or like term.
D.
In this option we do have the
, so it is a like term of 
E.
It is also a like term because it contains the
.
In summary the like terms are:
A.-
D.
E.
Answer:
Impossible
Step-by-step explanation:
In 5x^2-4x+3=0, standard form, substitute these values in the quadratic formula:
a = 5; b = -4; c = 3
The quadratic formula is 
(ignore the weird capital A)
Substitute a b and c:

Simplify:

Because
is the square root of a negative number, the answer would be imaginary.
Therefore, there are not solutions to this equation.
A solution is the same as the roots or zeroes, where the graph would cross the x-axis when graphed.
The graph never meets the x-axis. It looks like this: