Answer:
Choice A.
Step-by-step explanation:
Use graphing calculator
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
A point (-0.8, 0.6) will be a point on the unit circle in the second quadrant. Since it is a unit circle, its radius is 1, and we have
sin(α) = y = 0.6
cos(α) = x = -0.8
tan(α) = y/x = 0.6/-0.8 = -0.75
The angle is α = arccos(-0.8) ≈ 143.13°
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For the unit circle, the trig values are always the coordinates or their ratio as shown above, regardless of quadrant.
If ya distribute it it comes together as 24 + 8x soo i hope this helped
