Answer:
Owen can afford to keep and drive the car for 4 days.
Step-by-step explanation:
The total owing on the car rental is R(d) = ($77.25/day)d + ($0.12/mile)m ≤ $330. Substitute 1 for d (that is, the rental is for 1 day) and 175 for m:
R(d) = ($77.25/day)(d days) + ($0.12/mile)(175 miles) ≤ $330
= $77.25d + $21 ≤ $330
This simplifies to $77.25d + $21 ≤ $330, or
$77.25d ≤ $309
Solving this by dividing both sides of the above equation by $77.25, we get
d = ($309)/($77.25) = 4
4x+6=38
-6=-6
4x=32
4x/4=32/4
x=8
.........................p
Answer:
Any letter you want
Step-by-step explanation:
The letters used for variables don't matter. What matters is what the variables represent: their numerical value. Variables are just to help identify what needs solving. Other common variables are a, b, and c which are found frequently in trigonometry. To answer your question, there are no "next letters." You can use any letter you'd like as a variable because it holds the same numerical value. Basically, the whole alphabet is at your disposal.
3 miles. Using cross-multiplication, 3×4 = 12 and 4×1 = 4, so 12/4 = 3.
Answer: 334
Step-by-step explanation:
6 consecutive numbers can be written as:
n, n+1, n+2, n + 3, n + 4, n + 5,
The addition of those 6 numbers is:
n + n+1 + n+2 + n + 3 + n + 4 + n + 5
6n + 1 + 2 + 3 + 4 + 5 = 6n + 15
Let's find the maximum n possible:
6n + 15 = 2020
6n = 2020 - 15 = 2005
n = 2005/6 = 334.16
The fact that n is a rational number means that 2020 is can not be constructed by adding six consecutive numbers, but we know that with n = 334 we can find a number that is smaller than 2020, and with n = 335 we can found a number bigger than 2020.
So with n = 334 we can find one smaller.
6*334 + 15 = 2019
and we can do this for all the values of n between 1 and 334, this means that we have 334 numbers less than 2020 that can be written as a sum of six consecutive positive numbers.