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Travka [436]
3 years ago
10

Given: F3 = 5 k + 515 g F4 = 945k - 713 g ; find F20

Mathematics
1 answer:
Alisiya [41]3 years ago
8 0

9514 1404 393

Answer:

  F20 = 15985k -20361g

Step-by-step explanation:

If we assume F3 and F4 are the third and fourth terms of an arithmetic sequence, the common difference is ...

  d = F4 -F3 = (945k -713g) -(5k +515g)

  d = 940k -1228g

Then the 20th term is ...

  F20 = F3 +(20 -3)d = (5k +515g) +17(940k -1228g)

  F20 = 15985k -20361g

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Which value below is included in the solution set for the inequality statement?
mixas84 [53]

You haven't provided any value, but I can tell you the solution set for the inequality.

First of all, expand both sides:

-3x+12 > 6x-6

Add 3x to both sides:

12 > 9x-6

Add 6 to both sides:

18 > 9x

Which is of course equivalent to

9x < 18

Divide both sides by 9

x < 2

So, every number smaller than 2 is part of the solution of this inequality.

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3 years ago
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Montano1993 [528]

\frac{ {x}^{2}  - 7x + 12}{ {x}^{2} - x - 12 }  \\  =  \frac{ {x}^{2} - 4x - 3x + 12 }{ {x}^{2} - 4x + 3x - 12 }  \\  =  \frac{x(x - 4) - 3(x - 4)}{x(x - 4) + 3(x - 4)}  \\  =  \frac{(x - 3)(x - 4)}{(x + 3)(x - 4)}  \\  =  \frac{x - 3}{x + 3}

x² - x - 12 ≠ 0

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(B)

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3 years ago
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The second answer is D) it will increase 8 times
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Tan x-1/tan x+1=1-cot x/1+cotx
ikadub [295]
\dfrac{\tan x-1}{\tan x+1}=\dfrac{\frac{\sin x}{\cos x}-1}{\frac{\sin x}{\cos x}+1}=\dfrac{\sin x-\cos x}{\sin x+\cos x}=\dfrac{1-\frac{\cos x}{\sin x}}{1+\frac{\cos x}{\sin x}}=\dfrac{1-\cot x}{1+\cot x}
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