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:
Step-by-step explanation:
m has a slope of -2 and a y intercept of 4
y = -2x + 4
n has a slope of 1 and a y intercept of -1
y = 1x - 1
y = x - 1
Ah!... the invisible questions in a sideways posted image.
t has a slope of zero and a y intercept of 3
y = 0x + 3
y = 3
p has infinite slope and no y intercept.
x = -5
Answer:
0.9 , 0.088 , 0.032
Step-by-step explanation:
0.9 --> 1st greatest
0.088 --> 2nd greatest
0.032 --> 3rd greatest
So, 0.9 , 0.088 , 0.032
Answer:
10j^2 +j
Step-by-step explanation:
I added the first two terms since they have the same exponent and then left the j because its not squared or anything
