Show that 3 · 4^n + 51 is divisible by 3 and 9 for all positive integers n.

To prove that 3·4ⁿ + 51 is divisible by 3 and 9, we have;

3·4ⁿ is divisible by 3 and 51 is divisible by 3

Where we have;

=  3·4ⁿ + 51

= 3·4ⁿ⁺¹ + 51

- = 3·4ⁿ⁺¹ + 51 - (3·4ⁿ + 51) = 3·4ⁿ⁺¹ - 3·4ⁿ

- = 3( 4ⁿ⁺¹ - 4ⁿ) = 3×4ⁿ×(4 - 1) = 9×4ⁿ

∴ - is divisible by 9

Given that we have for S₀ =  3×4⁰ + 51 = 63 = 9×7

∴ S₀ is divisible by 9

Since   - is divisible by 9, we have;

- =   - is divisible by 9

Therefore is divisible by 9 and  is divisible by 9 for all positive integers n

Step-by-step explanation:

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl