11
(9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99)
Those numbers divisible by 9 are the multiple of 9; thus need to know how many multiples of 9 there are between 1 and 100:

100 ÷ 9 = 11 r 1 ÷ last multiple of 9 is 11 × 9 (= 99)
→ There are 11 - 1 + 1 = 11 numbers between 1 and 100 which are divisible by 9.

(They are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.)

7 0

3 years ago

(8+k³-6k⁴)+(4-3k⁴+8k²)

ANEK [815]

Answer:

12 + 8k² + k³ - 9k⁴

Step-by-step explanation:

(8 + k³ - 6k⁴) + (4 - 3k⁴ + 8k²)

8 + k³ - 6k⁴ + 4 - 3k⁴ + 8k²

(8 + 4) + (8k²) + (k³) + (-6k⁴ -3k⁴)

(12) + (8k²) + (k³) + (-9k⁴)

12 + 8k² + k³ - 9k⁴

6 0

3 years ago