gratis formulas matematicas enem 0609

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α α
𝑎
𝑠𝑒𝑛𝐴
= 
𝑏
𝑠𝑒𝑛�̂�
= 
𝑐
𝑠𝑒𝑛�̂�
𝐷. 𝑑
2
(𝐵 + 𝑏). ℎ
2
𝑏. ℎ
2
𝑎. 𝑏. 𝑠𝑒𝑛𝜃
2
𝜃: â𝑛𝑔𝑢𝑙𝑜 𝑓𝑜𝑟𝑚𝑎𝑑𝑜 𝑝𝑒𝑙𝑜𝑠 𝑙𝑎𝑑𝑜𝑠 𝑎 𝑒 𝑏
𝑐𝑎𝑡𝑒𝑡𝑜. 𝑐𝑎𝑡𝑒𝑡𝑜
2
𝐿2√3
4 𝐿√3
2
𝜋𝑅2
𝜋𝑅
𝜋𝑅2𝛼
360°
α
𝜋(𝑅2 − 𝑟2)
2𝜋𝑅ℎ 𝜋𝑅2
𝜋𝑅𝑔 𝜋𝑅2 𝐴𝐵. ℎ
3
𝐴𝐵. ℎ
3
4𝜋𝑅2 4
3
𝜋𝑅3
𝜃
𝑐𝑎𝑡𝑒𝑡𝑜 𝑜𝑝𝑜𝑠𝑡𝑜
ℎ𝑖𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑎
𝜃
𝑐𝑎𝑡𝑒𝑡𝑜 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡𝑒
ℎ𝑖𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑎
𝜃
𝑐𝑎𝑡𝑒𝑡𝑜 𝑜𝑝𝑜𝑠𝑡𝑜
𝑐𝑎𝑡𝑒𝑡𝑜 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡𝑒
𝜃 = 
1
𝑐𝑜𝑠𝜃
𝜃 = 
1
𝑠𝑒𝑛𝜃
𝜃 = 
1
𝑡𝑔𝜃
𝒔𝒆𝒏𝟐𝜽 + 𝒄𝒐𝒔𝟐𝜽 = 𝟏
{
−1 ≤ 𝑠𝑒𝑛𝜃 ≤ 1
−1 ≤ 𝑐𝑜𝑠𝜃 ≤ 1
𝑛ú𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑓𝑎𝑣𝑜𝑟á𝑣𝑒𝑖𝑠
𝑛ú𝑚𝑒𝑟𝑜 𝑑𝑒 𝑐𝑎𝑠𝑜𝑠 𝑝𝑜𝑠𝑠í𝑣𝑒𝑖𝑠
�̅� = 1 − 𝑃
(𝑎1+ 𝑎𝑛).𝑛
2
𝑎1(𝑞
𝑛−1)
𝑞−1
𝑆∞ = 
𝑎1
1 − 𝑞
log𝑏 𝑎 = 𝑐 ⇔ 𝑏
𝑐 = 𝑎
log𝑏 1 = 0
log𝑏 𝑏 = 1
log𝑏 𝑎 + log𝑏 𝑐 = log𝑏(𝑎. 𝑐)
log𝑏 𝑎 − log𝑏 𝑐 = log𝑏(
𝑎
𝑐
)
log𝑏𝑚 𝑎 log𝑏 𝑎
log𝑏 𝑎
𝑚 = 
1
𝑚
. log𝑏 𝑎
𝑏log𝑏 𝑎 = 𝑎
log𝑏 𝑎 = 
log𝑐 𝑎
log𝑐 𝑏