taylor cos(3t)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for partial fractions long division line

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Steps Examples

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complex form 8−√−246

Solution

8−√2√3 i

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Solution steps

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질량 보존의 법칙

질량 보존의 법칙은 어떤 원자도 화학 반응에서 생성되거나 파괴될 수 없다고 말한다. 반응물의 원자 수는 생성물의 원자 수와 균형을 이루어야 합니다.

방정식 양쪽에 있는 각 원소의 원자 수를 비교하고, 그 수가 같도록 계수를 변경한다

검사에 의한 반응의 균형

대수적으로 반응 균형

따라서 균형 잡힌 화학 반응은 다음과 같습니다:

P₄+5O₂⇒ 2P₂O₅

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

taylor cos(3t)

Solution

Solution steps

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