You must drive more than 40 miles to make option A the cheaper plan
<em><u>Solution:</u></em>
Two payment options to rent a car
Let "x" be the number of miles driven in one day
<em><u>You can pay $20 a day plus 25¢ a mile (Option A)</u></em>
25 cents is equal to 0.25 dollars
OPTION A : 20 + 0.25x
<em><u>You pay $10 a day plus 50¢ a mile (Option B)</u></em>
50 cents equal to 0.50 dollars
Option B: 10 + 0.50x
<em><u>For what amount of daily miles will option A be the cheaper plan ?</u></em>
For option A to be cheaper, Option A must be less than option B
Option A < Option B

Solve the inequality
Add -0.50x on both sides

Add - 20 on both sides,



Divide both sides by 0.25

Thus you must drive more than 40 miles to make option A the cheaper plan
10 gallons= 208
1 gallon: 208 ÷10= 20.8
(when you divide by 10, the decimal place jumps front 1 place!)
hope this helps!

One really simple way to find an equivalent fraction is just to multiply the numerator and denominator.
For example, for the first answer,
, I simply multiplied the numerator and denominator each by 2. This gives you an equivalent fraction.
You can then continue with this process using numbers such as 3, 4, 5, and 6 as your multiplication factor.
Answer:
1526.25 so in one day the interest will have added 26.25
Step-by-step explanation:
The solution of the equation which is as described in the task content when expressed in its simplest form is; -5/13.
<h3>What is the solution of the equation as described in the task content?</h3>
It follows from the task content that the solution to the equation is to be determined and expressed in its simplest form.
Given;
-1/2 -1/2(4/5x+1) = -2 -3x
By distributing the coefficients: we have;
-1/2 - 2/5x - 1/2 = -2 -3x
Hence, by collecting like terms; we have;
-1/2 - 1/2 + 2 = -3x + (2/5)x
1 = -13x/5
Therefore; by cross-multiplication; we have;
5 = -13x
Divide both sides by; -13.
x = -5/13.
Ultimately, the solution of the equation which is as described in the task content when expressed in its simplest form is; -5/13.
Read more on solution of an equation;
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