Triangle area formula: base*height/2
so, 14*10/2 = 7*10 = 70 sq. in.
(f o g)(x)
g(x) = x + 2
Since it’s telling you to use g(x), you use (x + 2) as your new x value for function f.
f(x + 2) = 5(x + 2) + 5
= 5x + 10 + 5
= 5x + 15
If a = first term and r = common ratio we have
a + ar + ar^2 = 13 and ar^2 / a = r^2 = 9
so r = 3
and a + 3a + 9a = 13
so a = 1
so they are 1,3 and 9
2.
in geometric series we have
4 , 4r ,4r^2 , 60
Arithmetic;
4, 4r , 4r + d , 4r + 2d
so we have the system of equations
4r + 2d = 60
4r^2 = 4r + d
From first equation
2r + d = 30
so d = 30 - 2r
Substitute for d in second equation:-
4r^2 - 4r - (30-2r) = 0
4r^2 - 2r - 30 =0
2r^2 - r - 15 = 0
(r - 3)(2r + 5) = 0
r = 3 or -2.5
r must be positive so its = 3
and d = 30 - 2(3) = 24
and the numbers are 4*3 = 12 , 4*3^2 = 36
first 3 are 4 , 12 and 36 ( in geometric)
and last 3 are 12, 36 and 60 ( in arithmetic)
The 2 numbers we ause are 12 and 36.
The HA Theorem is a special case of the AAS postulate.
