Answer:
Cost to buy and run the first car for 5 years=$24000
Cost to buy and run the second car for 5 years=$26000
I think we should buy the first car, because It is cheaper to buy and maintain
Step-by-step explanation:
first car:
Cost $18000 to buy
Cost $100 to maintain each month
5years =5 x 12=60months
Cost to maintain for 60months=100x60=$6000
First car cost to buy and run over the 5 year period=18000+6000=$24000
Second car:
It cost $22400 to buy
Cost $60 to maintain each month
5years=60months
Cost to maintain for 60months is 60 x 60=$3600
Cost to buy and run over the second car for 5years is 22400+3600=$26000
83,034.6 - 82,803.6 = 231 miles
231/16.5 = 14 miles per gallon
Answer:
The 4th one is the least so you would put that as the first one, then the 2nd one, then the 1st one, then the 3rd one.
Step-by-step explanation:
The fourth one is first because 10 to the power of 0 is 0 times 10 to the power of 1 is still 0 minus 1 to the power of 10 is 1 so 0 minus 1 is -1. The second one is second because 10 to the power of 0 times 10 to the power of 1 is 0, times 1 to the power of 10 is still 0. The first one is third in line because 10 to the power of 0 plus 10 to the power of 1 is 10 times 1 to the power of 10 is 10. The third one is last because its the greatist, 10 to the power of 0 plus 10 to the power of 1 is 10 plus 1 to the power of 10 is 11.
Answer:
p = 3, q = 2.5
Step-by-step explanation:
Given that x = - 4 is a root of the equation, then
(- 4)² - 4p - 4 = 0, that is
16 - 4p - 4 = 0
12 - 4p = 0 ( subtract 12 from both sides )
- 4p = - 12 ( divide both sides by - 4 )
p = 3
Also given that the equation x² + px + q = 0 has equal roots
Thus the value of the discriminant b² - 4ac = 0
Here a = 1, b = p and c = q, thus
p² - 4q = 0 ← substitute p = 3 from above into the equation
9 - 4q = 0 ( subtract 9 from both sides )
- 4q = - 9 ( divide both sides by - 4 )
q = 2.5