The vertex (minimum) of the quadratic ax² +bx +c is located at x=-b/(2a). This means the minimum value of f(x) will be found at x = -3/(2*1) = -1.5.
Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.
On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057
Answer: Choice C
7 divided by cosec 50 degrees
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Explanation:
Abbreviations:
sec = secant
cot = cotangent
cosec = cosecant (csc is also a widely used abbreviation)
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Let the horizontal leg be x
This leg is opposite the angle 50 degrees. The hypotenuse is 7
Opposite = x
Hypotenuse = 7
We'll either use the sine rule or the cosecant (csc) rule
sine = opposite/hypotenuse
cosecant = hypotenuse/opposite
Since "sine" is nowhere to be found in the answer choices, we'll use the cosecant rule
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cosecant = hypotenuse/opposite
csc(angle) = hypotenuse/opposite
csc(50) = 7/x
x*csc(50) = 7
x = 7/csc(50)
This is why the answer is choice C
Answer:
12
Step-by-step explanation:
Rotation doesn't alter the lengths.
6×2 = 12
Just do 47+118 and it equals 165. Jorge started with 165 tickets.
Well, if we see the graph of x^2
the vertex is (0,0)
if we reflect it acrosss the x axis, teh vertex is still (0,0)
verteical strech doesn't change the vertex
vertex of g(x) is same as f(x)
first answer