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7nadin3 [17]
3 years ago
10

HELP ME PLZZZZZZZZZZZZZZZZZ!!!!!!!!!!

Mathematics
1 answer:
aev [14]3 years ago
3 0

Answer:

45

Step-by-step explanation:

I multiplied 7 x 9 and got 63 so I figure 5 x 9 is right.

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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
Somebody link me a video on how to do proofs
Artyom0805 [142]
Https://youtu.be/fSu1LKnhM5Q
7 0
2 years ago
Surds and roots<br> look at picture
FinnZ [79.3K]

\huge \boxed{\mathfrak{Question} \downarrow}

  • Expand & simplify ⇨ ( \sqrt{10}  -  \sqrt{2} ) ^{2}. Give your answer in the form b - c \:  \sqrt{5} where b & c are integers.

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

( \sqrt { 10 } - \sqrt { 2 } ) ^ { 2 }

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{10}-\sqrt{2}\right)^{2}.

\left(\sqrt{10}\right)^{2}-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}

The square of \sqrt{10} is 10.

10-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}

Factor 10=2\times 5. Rewrite the square root of the product \sqrt{2\times 5} as the product of square roots \sqrt{2}\sqrt{5}.

10-2\sqrt{2}\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}

Multiply \sqrt{2} and \sqrt{2} to get 2.

10-2\times 2\sqrt{5}+\left(\sqrt{2}\right)^{2}

Multiply -2 and 2 to get -4.

10-4\sqrt{5}+\left(\sqrt{2}\right)^{2}

The square of \sqrt{2} is 2.

10-4\sqrt{5}+2

Add 10 and 2 to get 12.

\boxed{ \boxed{\bf\:12-4\sqrt{5} }}

  • Here, b & c are integers where \boxed{ \sf \: b = 12 \: and \: c = 4}
7 0
2 years ago
Read 2 more answers
Use the distributive property to remove the parentheses.<br><br> -8(3w - u - 4)
N76 [4]
-8(3w-u-4)
-24w+8u+32
when multiplying two negatives you get a posative
5 0
3 years ago
Which function is equivalent to y=2(x+3)2+5?
Pepsi [2]

Answer:

y=2x2+12x+23

Step-by-step explanation:

2×2+12x+23

2(2+6x)+23

2×2(1+3x)+23

2(1+3x)2+23

I hope it will help you.

8 0
2 years ago
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