Answer:
The figure above shows the graph of a function f with domain 0 ≤ x ≤ 4. Which of the following statements are true? Show Video ... If there is no c, where -2 < c < 2, for which f'(c) = 0, which of the following statements must be true? Show Video ...
Step-by-step explanation:
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Answer:
Step-by-step explanation:
If a = 0, f(a) = f(0) = 0 + 0 + 7 = 7;
If a = 1, f(1) = 3(1^2) + 2(1) + 7 = 12;
and so on. Normally an 'a' value would be given as part of this problem.
Actually, the value of f(a) is given and is f(a) = 3a^2+2a+7 for input value a.
Answer:
Step-by-step explanation:
Given the equation
Sin(5x) = ½
5x = arcSin(½)
5x = 30°
Then,
The general formula for sin is
5θ = n180 + (-1)ⁿθ
Divide through by 5
θ = n•36 + (-1)ⁿ30/5
θ = 36n + (-1)ⁿ6
The range of the solution is
0<θ<2π I.e 0<θ<360
First solution
When n = 0
θ = 36n + (-1)ⁿθ
θ = 0×36 + (-1)^0×6
θ = 6°
When n = 1
θ = 36n + (-1)ⁿ6
θ = 36-6
θ = 30°
When n = 2
θ = 36n + (-1)ⁿ6
θ = 36×2 + 6
θ = 78°
When n =3
θ = 36n + (-1)ⁿ6
θ = 36×3 - 6
θ = 102°
When n=4
θ = 36n + (-1)ⁿ6
θ = 36×4 + 6
θ = 150
When n =5
θ = 36n + (-1)ⁿ6
θ = 36×5 - 6
θ = 174°
When n = 6
θ = 36n+ (-1)ⁿ6
θ = 36×6 + 6
θ = 222°
When n = 7
θ = 36n + (-1)ⁿ6
θ = 36×7 - 6
θ = 246°
When n =8
θ = 36n + (-1)ⁿ6
θ = 36×8 + 6
θ = 294°
When n =9
θ = 36n + (-1)ⁿ6
θ = 36×9 - 6
θ = 318°
When n =10
θ = 36n + (-1)ⁿ6
θ = 36×10 + 6
θ = 366°
When n = 10 is out of range of θ
Then, the solution is from n =0 to n=9
So the equation have 10 solutions in the range 0<θ<2π