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katrin2010 [14]
4 years ago
11

A number to the 11th power divided by the same number to the 8th power equals 27. what is the number?

Mathematics
2 answers:
valkas [14]4 years ago
8 0
The answer is 3.
If n^11/n^8=27 then, n^3=27.
The cube root of 27 is 3.
So the answer is 3.
Lubov Fominskaja [6]4 years ago
6 0
The answer to this question iis3
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If you know that​ R' (20)=50 and​ R''(20)=-4, approximate​ R'(24)
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Then

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jek_recluse [69]

f(x) =  {x}^{2}  - 3x + 2

<h2>••••••••••••••••••••••••••••••••••••••••</h2>

put x = (-1)

f( - 1) =  {( - 1)}^{2}  - 3 \times ( - 1) + 2  \\  \\   1  + 3 + 2  \\  \\   6.

<h2>••••••••••••••••••••••••••••••••••••••••</h2>

put x = (z)

f(z) =  {z}^{2}  - 3 \times z + 2  \\  \\  {z}^{2}  - 3z + 2  \\  \\  {z}^{2} - 2z - z + 2 \\  \\ z(z - 2) - 1(z - 2) \\  \\ (z - 1)( z -2 ) \\  \\ (z - 1) = 0  \:  \: , \:  \: (z - 2) = 0\\  \\ z =  1 \:,  \: 2

<h2>••••••••••••••••••••••••••••••••••••••••</h2>

put x = (x+1)

f(x + 1)  = {x}^{2}  -  3x + 2 \\  \\  {(x + 1)}^{2}  - 3 (x + 1) + 2 \\  \\  {x}^{2}  + 1 + 2x - 3x - 3 + 2 \\  \\  {x}^{2}  - x \\  \\ x(x - 1) \\  \\ x(x - 1) = 0   \\  \\  x - 1 =  \frac{0}{x}  \\  \\ x - 1 = 0 \\  \\ x = 1.

<h2>•••••••••••••••••••••••••••••••••</h2>

put x = (√2 + 1 )

f( \sqrt{2}  + 1) =  {x}^{2}  - 3x + 2 \\  \\  {( \sqrt{2} + 1) }^{2}  - 3( \sqrt{2}  + 1) + 2 \\  \\ 2 + 1 +  2\sqrt{2}  - 3 \sqrt{2}  - 3 + 2 \\  \\ 3 - 1 \sqrt{2}  - 1 \\  \\ 2 -  \sqrt{2}  \\   \\  2 - 1.4 (optional \:  \:  \: steps)\\  \\ 0.6 \:  \:

4 0
2 years ago
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