Find all real zeros of the function. f(x)=4(x^2-1)(x-3)(x+3)^2
1 answer:
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Answer and Step-by-step explanation:
<u>Trigonometric Function (These only work for right triangles):</u>
SOH-CAH-TOA
S = Sine = sin
C = Cosine = cos
T = Tangent = tan
O = Opposite (side)
H = Hypotenuse (side)
A = Adjacent (side)
SOH = sin(angle) =
CAH = cos(angle) =
TOA = tan(angle) =
1. sinQ (Q is the angle) =
Use Pythagorean Theorem to find side PQ.
2. cosQ =
3. tanQ =
Answer:
1) 36
b) 5
c) 3.0
Step-by-step explanation:
1) The recursive formula that defines the given sequence is
That means we keep adding 4 to the subsequent terms:
The sequence will be:
12,16,20,24,28,32,36,...
Therefore the seventh term is 36.
2) The sequence is recursively defined by;
This means, we have to keep subtracting 5 from the subsequent terms.
The sequence will be;
20,15,10,5,...
Therefore the fourth term is 5
3) The sequence is recursively defined by:
f(n+1)=f(n)+0.5
where f(1)=-1.5
This means that, the subsequent terms can be found by adding 0.5 to the previous terms.
The sequence will be:
-1.5,-1.0,-0.5,0,0.5,1,1.5,2.0,2.5,3.0,....
Therefore f(10)=3.0