skalarprojektion (2, 4), (-1, 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 inverse tangent line

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scalar projection (2, 4), (−1, 5)

Solution

9√55

Decimal

4.02492…

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Solution steps

One step at a time

Finding the scalar projection of ⃗ b onto ⃗ a : comp_⃗ a(⃗ b )=⃗ b ·⃗ a |⃗ a |

⃗ a =(

), ⃗ b =(

−1

)

|⃗ a |: |(

)|=2√5

⃗ b · ⃗ a : (

−1

)(

)=18

⃗ b · ⃗ a |⃗ a | =182√5

Simplify

=9√55

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Notebook

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

skalarprojektion (2, 4), (-1, 5)

Solution

Solution steps

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