f(x)=sqrt((1+(30e^{-0.3x))^2)}, f(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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symmetry 6a²+5a+10−(4a²+6a+12)

Solution

a=14

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Axis of Symmetry

For a parabola in standard form y=ax²+bx+c the axis of symmetry is the vertical line that goes through the vertex x=−b2a

y=6a²+5a+10−(4a²+6a+12)

Rewrite in standard form: y=2a²−a−2

y=2a²−a−2

Axis of Symmetry for y=ax²+bx+c is x=−b2a
a=2, b=−1

a=−(−1)2· 2

Refine

a=14

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

f(x)=sqrt((1+(30e^{-0.3x))^2)}, f(5)

Solution

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