Answer:
The first term is:
Step-by-step explanation:
The arithmetic sequence is defined by
where
is the first term
is the common difference
as
the 17th term of an arithmetic sequence is 51.
i.e.
so
∵
,
,
∵
Therefore, the first term is:
The derivative, f'(x) = 6x^2+1, is never negative, so f(x) is monotonic, hence invertible.
f'(-2) = 6(-2)²+1 = 25
If point (-2, -26) is on the graph of function f(x), and the slope is 25 there, then (-26, -2) is on the graph of f⁻¹(x), and the slope is 1/25 there. The equation of the tangent line throught that point can be written in point-slope form as
... y +2 = (1/25)(x +26)
Answer:
Step-by-step explanation:
Here you are require to find the equation of the hyperbola given that the center (h,k), the coordinates of the vertices and those of the co-vertices can be determined from the diagram given
The sharp turning points of the curves give the vertices at (-4,2) and (6,2)
Joining the vertices with a straight line will form the transverse axis with length 2a .
To find the length of the transverse axis 2a will be ; 6--4=10. 2a=10 hence a=10/2 =5
a=5
Find the center of the hyperbola at (h,k) by finding the intersecting point of the diagonals of the rectangle in the diagram
The center identified will be (h,k) = (1,2)
To find the length of the conjugate axis 2b will be ; the length between points (1,4) and (1,0) which are the coordinates of the co-vertices in the hyperbola. 2b= 4-0=4 , b=4/2 = 2
b=2
The standard equation of the hyperbola with center (h,k) is written as ;
where (h,k) is center of hyperbola, (h±a,k) is coordinate of the vertices and (h,k±b) are coordinates of co-vertices.
Substitute values of a, b, h, and k in equation as