Answer:
The answer for your question is "False"
First we need to find the total annual costs for both the plans.
In case of leasing:
Fixed monthly cost = $420
So, yearly cost = 420 x 12 = $5040
Cost per mile = $0.08
For x miles driven, the cost per year for leasing will be = 5040 + 0.08x
In case of purchasing:
Fixed yearly cost = $4600
Cost per mile = 0.10
For x miles driven, the cost per year for leasing will be = 4600 + 0.10x
We want to find for what number of miles will the cost of leasing will be no more expensive than the cost of purchasing.
So,
Cost of leasing ≤ Cost of purchasing
5040 + 0.08x ≤ 4600 + 0.10x
440 ≤ 0.02x
22000 ≤ x
Thus, if if the number of miles driven are equal to or less than 22,000 leasing will be no more expensive than purchasing.
Answer:
see explanation
Step-by-step explanation:
Equation for Fast Internet is
45 + 50.45x
Equation for Quick Internet is
57.95x
To find when they cost the same , equate the 2 equations
45 + 50.45x = 57.95x ← required equation
Solve the equation by subtracting 50.45x from both sides
45 = 7.5x ( divide both sides by 7.5 )
6 = x
Both services would cost the same after 6 months
Hi! I will try my best to answer your question, the wording provided is unclear.
If the 10% of the original cost of something is $1.50, then the total cost is $15.00.
Assuming this is the discounted price, which would be 90% of the original cost, the original cost would be $16.67.
To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
