4secx+5=13 subtract 5 from both sides
4secx = 8 divide both sides by 4
secx = 2
If the secx = 2, then cosx = 1/2
And this happens at x = pi/3 and x = 5pi/3
I suppose that you're technically correct......the common difference is -5 [ we are subtracting 5 each time ]
Good idea.....!!!
The last term is 253
And the common difference in terms is 5
So......the 20th term from the last =
253 - 5(n - 1) =
253 - 5(20 - 1) =
253 - 5(19) =
158
( [.9 + .8 + .7 + .6] - [.4 + .3 + .2 + .1 ] ) * .5 =
(3 - 1 ) * .5 =
2 * . 5 =
1
(0 + 3) (10 + 3) =
(30 + 9)(20 + 3)=
(600 + 180 + 90 + 27) (30 + 3) =
(..............+ 81)(40 + 3)....... etc.
So (3 x 13 x 23 x........183 x 193) / 5 will have a remainder =
3^20 mod 5 =
(3^4)^5 mod 5 =
(3^4) mod 5 =
a) The length of the secant RT, times its external segment, ST, equals the square of the tangent segment TU.
b) RT * ST = TU^2
9 * 4 = TU^2
36 = TU^2 take the square root of both sides
6 = TU
6 - 2x = 6x - 10x + 20 simplify
6 - 2x = -4x + 20 add 4x to both sides, subtract 6 from both sides
2x = 14 divide by 2
x = 7
f(-9) =
(-9)^2 + 5 =
81 + 5 =
86
95 : 114 =
5 * 19 : 6 * 19 divide by 19
5 : 6
