Answer:
12 and 13
Step-by-step explanation:
Answer:
<h2>3(cos 336 + i sin 336)</h2>
Step-by-step explanation:
Fifth root of 243 = 3,
Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),
then r^5( cos Ф + i sin Ф )^5 = 243(cos 240 + i sin 240).
Equating equal parts and using de Moivre's theorem:
r^5 =243 and cos 5Ф + i sin 5Ф = cos 240 + i sin 240
r = 3 and 5Ф = 240 +360p so Ф = 48 + 72p
So Ф = 48, 120, 192, 264, 336 for 48 ≤ Ф < 360
So there are 5 distinct solutions given by:
3(cos 48 + i sin 48),
3(cos 120 + i sin 120),
3(cos 192 + i sin 192),
3(cos 264 + i sin 264),
3(cos 336 + i sin 336)
Answer:
Step-by-step explanation:
nth term = a +(n-1)d
a3 = 116 ; a + 2d = 116 ---------(i)
a7 = 180; a + 6d = 180 -------(ii)
multiply (ii) by -1. so a will be eliminated
a + 2d = 116 ---------(i)
(ii)*-1 <u>-a - 6d = -180</u> -------(ii) { Now add the two equations}
- 4d = -64
d = -64/-4
d = 16
Plug in the value of d in equation (i),
a + 2*16 = 116
a + 32 = 116
a = 116 - 32
a = 84
12th term = 84 + 11* 16 = 84 + 176 = 260
Philip equation
ordered data: 3o(oranges), 3g(green), 6y, 8r, 13b(black, 15b(blue
mean: 8 mode: 3 range: 12
Answer:
an = 2.5n -0.2
Step-by-step explanation:
an = 2.5n -0.2
