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gladu [14]
3 years ago
14

(r - 3)(3r + 6)= ? .....

Mathematics
1 answer:
irga5000 [103]3 years ago
4 0
R(3r+6)-3(3r+6)
3r² +6r-9r-18
3r²-3r-18
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Find the equation of the normal of the curve at the given points :D thankiewss
Ahat [919]
We want to find the equation of the normal line of y=\dfrac{20-x}{3x} at the point P=(x_0,y_0), where y_0=3. First calculate x_0. We have:

y_0=f(x_0)=\dfrac{20-x_0}{3x_0}\\\\\\3=\dfrac{20-x_0}{3x_0}\quad|\cdot3x_0\\\\\\3\cdot3x_0=20-x_0\\\\9x_0=20-x_0\\\\9x_0+x_0=20\\\\10x_0=20\quad|:2\\\\\boxed{x_0=2}

Now, when we know that P=(x_0,y_0)=(2,3) we can write an equation of the normal line as:

\boxed{y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)}

Calculate f'(x_0):

f'(x)=\left(\dfrac{20-x}{3x}\right)'=\left(\dfrac{20}{3x}\right)'-\left(\dfrac{x}{3x}\right)'=\dfrac{20}{3}\cdot\left(\dfrac{1}{x}\right)'-\left(\dfrac{1}{3}\right)'=\\\\\\=\dfrac{20}{3}\cdot\left(-\dfrac{1}{x^2}\right)-0=\boxed{-\dfrac{20}{3x^2}}\\\\\\\\f'(x_0)=f'(2)=-\dfrac{20}{3\cdot2^2}=-\dfrac{20}{3\cdot4}=-\dfrac{20}{12}=\boxed{-\frac{5}{3}}

and the equation of the normal line:

y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\y-3=-\dfrac{1}{-\frac{5}{3}}(x-2)\\\\\\y-3=\dfrac{3}{5}(x-2)\\\\\\y-3=\dfrac{3}{5}x-\dfrac{6}{5}\\\\\\y=\dfrac{3}{5}x-\dfrac{6}{5}+3\\\\\\\boxed{y=\dfrac{3}{5}x+\dfrac{9}{5}}
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3 years ago
What is the value of c
agasfer [191]

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if im not mistaken its 121

Step-by-step explanation:

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AlekseyPX

Answer:

The sequence is recursive and can be represented by the function f(n + 1)= f(n) + 3/8

Step-by-step explanation:

i got it right on the assignment

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3 years ago
Im bored lemme get yallz IG's please dont report :) oh happy saint patricks day :)
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Answer:7+7

Step-by-step explanation:

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What is the answer if Mr Rowley has 16 homework papers and 14 exit tickets
Zanzabum

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30 I'm guessing you mean add them.

Step-by-step explanation:

16+14=30

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3 years ago
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