sec ^6 A−tan ^6A=(1+tan ^2A−tan ^2 A)(1+tan ^4
A+2tan ^2 A+tan ^2 A+tan ^4 A+tan ^4 A)
sec ^6 A−tan ^6 A=1+3tan ^2 A+3tan ^4A
L.H.S. = R.H.S.
Hence proved
Step-by-step explanation:
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How do i solve for x??
1 answer:
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Answer:
For a general equation:
y = f(x) = a*x^2 + b*x + c
The x-intercepts are given by:
And the vertex is the point (h, k) such that:
h = -b/2*a
k = f(h)
Part A)
We have the function f(x) = -16*x^2 + 22*x + 3
The x-intercepts are then:
The two intercepts are:
x = (-22 - 26)/(-32) = 1.5
x = (-22 + 26)/(-32) = -0.125
Part B:
The vertex is (h, k), such that:
h = -22/(2*(-16)) = -22/-32 = 0.6875
k = f(0.6875) = -16*(0.6875)^2 + 22*0.6875 + 3 = 10.6525
Then the vertex is: ( 0.6875, 10.6525)
If the leading coefficient is positive, then the vertex is a minimum
If the leading coefficient is negative, then the vertex is a maximum.
In this case the leading coefficient is -16, then we can conclude that the vertex is a maximum.
Part C:
To graph the function, we can graph the points that we already know (the vertex and the two x-intercepts) and connect them with a curve.
You could also add another few points so you have a guide to draw the curve, for example the point:
y = f(0) = -16*0^2 + 22*0 + 3
f(0) = 3
(0, 3)
