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.)
Answer:
sin²2θ. (cos θ sin θ). cos 2θ
Step-by-step explanation:
finding g'(x)
g'(x)
= 4 (cosθsinθ)³ . { cosθ. (sinθ)' + sinθ. (cosθ)' }
- (cosθ)' = - sinθ
- (sinθ)' = cosθ
= 4 (cosθsinθ)³ { cosθ. cos θ + sinθ.(-sin θ)}
= 4 (cosθsinθ)³{ cos²θ - sin²θ}
- cos²θ - sin²θ = cos 2θ
- 2sinθ cosθ = sin 2θ
= (4 cosθ sinθ)². (cosθ sinθ). { cos²θ - sin²θ}
= <u>sin²2θ. (cos θ sin θ). cos 2θ</u>
Answer:B
Step-by-step explanation:
Answer:
Step-by-step explanation:
40080 x 10
= 400800
