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2 years ago

12

Simplify the expression

-%208p%20%7D%5E%7B2%7D%20%29" id="TexFormula1" title=" {9p}^{4} \: x \: ( { - 8p }^{2} )" alt=" {9p}^{4} \: x \: ( { - 8p }^{2} )" align="absmiddle" class="latex-formula">

1 answer:

lions [1.4K]2 years ago

5 0

Answer:

-72p6x

Step-by-step explanation:

From <em>my </em>previous answer to a question similar

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Solve the system of equations using elimination<br> x-y=3<br> 2x+2y=50

pishuonlain [190]

Answer:

X=14

Y=11

Step-by-step explanation:

14-11=3

14x2=28

11x2=22

22+28=50

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2 years ago

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26. -2, 3, 8, 13, 18,...

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Step-by-step explanation:

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3 years ago

How do you write 5x - 4y = 1 in slope intercept form

Alexxandr [17]

First move the 4y to the right and the 1 to the left:

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3 years ago

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Find the original price of a pair of shoes if the sale price is $40 after a 60% discount

Arturiano [62]

Answer:

<em>Thus, the original price of the pair of shoes was $100.</em>

Step-by-step explanation:

<u>Percentages</u>

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Verify applying 60% discount:

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3 0

3 years ago

A) Use the limit definition of derivatives to find f’(x)

Ann [662]

<h3>1)</h3>

$\text{Given that,}\\\\f(x) = \dfrac{ 1}{3x-2}\\\\\text{First principle of derivatives,}\\\\f'(x) = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{ h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3(x+h) - 2} - \tfrac 1{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3x+3h -2} - \tfrac{1}{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{3x-2-3x-3h+2}{(3x+3h-2)(3x-2)}}{h}\\\\\\$

$~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{-3h}{(3x+3h-2)(3x-2)}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{-3h}{h(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \lim \limits_{h \to 0} \dfrac{1}{(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \cdot \dfrac{1}{(3x+0-2)(3x-2)}\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)(3x-2)}\\\\\\~~~~~~~=-\dfrac{3}{(3x-2)^2}$

<h3>2)</h3>

$\text{Given that,}~\\\\f(x) = \dfrac{1}{3x-2}\\\\\textbf{Power rule:}\\\\\dfrac{d}{dx}(x^n) = nx^{n-1}\\\\\textbf{Chain rule:}\\\\\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\\\\\text{Now,}\\\\f'(x) = \dfrac{d}{dx} f(x)\\\\\\~~~~~~~~=\dfrac{d}{dx} \left( \dfrac 1{3x-2} \right)\\\\\\~~~~~~~~=\dfrac{d}{dx} (3x-2)^{-1}\\\\\\~~~~~~~~=-(3x-2)^{-1-1} \cdot \dfrac{d}{dx}(3x-2)\\\\\\~~~~~~~~=-(3x-2)^{-2} \cdot 3\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)^2}$

8 0

2 years ago