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
Answer:
Hard questions
Step-by-step explanation:
Step-by-step explanation:
"identity" means an operation that does nothing.
For adding numbers, adding 0 changes nothing, so 0 is the identity for addition.
For multiplication, multiplying by 1 changes nothing, so 1 is the identity for multiplication.
66 is the answer
76+ 38= 114
114+ 66 = 180
So here are your variables:
# of adult tickets= x
# of student tickets 2x
so x+2x= 246 and solve
3x=246
x=82 adult tickets were sold
