log_{2}(2^5)

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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 expand factor rationalize

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Steps Examples

Generated by AI

vetores próprios (

1+	+2+
3+	+4

)

Solution

(

−3+√33

), (

−√33−3

)

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Solution steps

One step at a time

(

1+	2+
3+	4

)

Vetores próprios de uma matriz

Seja A uma matriz quadrada, η um vetor e λ um escalar que satisfaça Aη=λη,
então η é um vetor próprio de A e λ é o valor próprio associado a ele

Para encontrar os vetores próprios η, resolver (A−λ I)η=0 para cada valor próprio λ

Encontrar os valores próprios de (

1+	+2+
3+	+4

): λ=5+√332 , λ=5−√332

Vetores próprios de λ=5+√332 : (

−3+√33

)

Vetores próprios de λ=5−√332 : (

−√33−3

)

Os vetores próprios de (

1+	+2+
3+	+4

)

−3+√33

), (

−√33−3

)

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Notebook

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

log_{2}(2^5)

Solution

Solution steps

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