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pendenza 3x+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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autovalori (

0	9	3
2	0	4
3	7	0

)

Solution

λ=−5, λ=5+√1452 , λ=5−√1452

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Solution steps

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(

0	9	3
2	0	4
3	7	0

)

Autovalori della matrice

Sia A una matrice quadrata, η un vettore e λ uno scalare tale che Aη=λη,
allora λ è chiamato l'autovalore associato al corrispondente autovettore η di A

Gli autovalori di A sono le radici dell' equazione caratteristica det(A−λ I)=0

det((

0	9	3
2	0	4
3	7	0

)−λ(

1	0	0
0	1	0
0	0	1

)): −λ³+55λ+150

Risolvi −λ³+55λ+150=0: λ=−5, λ=5+√1452 , λ=5−√1452

Gli autovalori sono:

λ=−5, λ=5+√1452 , λ=5−√1452

Zero Exponent Rule Examples

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0	9	3
2	0	4
3	7	0

) oppure scegli qui sotto

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Notebook

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

pendenza 3x+5

Solution

Solution steps

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