Answer:
eeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
eeeeeeeeeeeeeeeeeeeeeeStep-by-step explanation:
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: It has no solution.
Step-by-step explanation:
3x - 12 - 4 = 3x + 4 - 4
3x - 8 = 3x
3x - 8 - 8 = 3x - 8
we can't subtract 3x-8
Doesn't add up; they need to have one solution or more for it to work.
Answer:
You have your x and y mixed up
Step-by-step explanation:
Look at your table. You labeled y1 and y2 where the x's are. y2 = 8, y1 = 4, x2 = 12 and x1 = 6. The slope then is (8-4)/(12-6) which is 4/6 or 2/3. You did run over rise instead of rise over run. Be careful. I see that a lot when I teach the concept to beginners.
