Mr johnson borrowed 750 for 1 year. He has to pay 6 percent simpe interest. How much interest will he pay?

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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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∂ ∂ x (√x¹¹−y²⁰)

Solution

11x¹⁰2√x¹¹−y²⁰

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Solution steps

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∂ ∂ x (√x¹¹−y²⁰)

Treat y as a constant

Apply the chain rule: 12√x¹¹−y²⁰ ∂ ∂ x (x¹¹−y²⁰)

=12√x¹¹−y²⁰ ∂ ∂ x (x¹¹−y²⁰)

∂ ∂ x (x¹¹−y²⁰)=11x¹⁰

=12√x¹¹−y²⁰ · 11x¹⁰

Simplify 12√x¹¹−y²⁰ · 11x¹⁰: 11x¹⁰2√x¹¹−y²⁰

=11x¹⁰2√x¹¹−y²⁰

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

Mr johnson borrowed 750 for 1 year. He has to pay 6 percent simpe interest. How much interest will he pay?

Solution

Solution steps

Number Line