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<h3>4 + 3s + 2 </h3>
→ Terms - <u>Three</u> (<u>Trinomial</u>)
→ Coefficient - <u>3</u>
→ Constants - <u>4</u> and <u>2</u>
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Answer:
Let A1=a1+a2+a3, A2=a2+a3+a4, and so on, A10=a10+a1+a2. Then A1+A2+⋯+A10=3(a1+a2+⋯+a10)=(3)(55)=165, so some Ai≥165/10=16.5, so some Ai≥17.
Step-by-step explanation:
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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>
