Negative, You'll have to explain the why part though,
Answer:
C) 0 ≤ x ≤ 25
Step-by-step explanation:
We are supposed to find a reasonable constraint so that the function is at least 300 i.e. the value of x at which f(x) is greater or equal to 300
A)x ≥ 0
Refer the graph
At x = 0
f(x)=300
On increasing the value of x , f(x) increases but at x = 12 it starts decreasing
So, x ≥ 0 can also have f(x)<300
So, Option A is wrong
B)−5 ≤ x ≤ 30
At x = -5
f(x) = 100
So, Option B is wrong since we require f(x) is greater or equal to 300
c)0 ≤ x ≤ 25
At x = 0
f(x)=300
At x = 12 , it starts decreasing
At x = 25
f(x)=300
So, The value of f(x) is at least 300 when 0 ≤ x ≤ 25
D)All real numbers
At x = 30
f(x)=0
But we require f(x) greater or equal to 300
Hence Option C is true
First off, we'll convert the mixed fraction to "improper" and check about, keeping in mind that, there are 5280 feet in 1mile.

90 days.
For this question you just need to find the least common multiple of 18 and 30. That can be found by first finding the prime factorization of each number. 18 = 2 * 3 * 3 and 30 = 2 * 3 * 5. Then you multiply the 2 * 3 * 3 * 5, you ignore the 3 from 30 because 18 has more 3s. The answer is 90.