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zaharov [31]
3 years ago
6

What is the value of n in the equation 0=18+128n^2?

Mathematics
1 answer:
lbvjy [14]3 years ago
7 0

Answer:

\large\boxed{\text{no soltion in the set of real numbers}}\\\boxed{n=-\dfrac{3}{8}i\ \vee\ n=\dfrac{3}{8}i\ \text{in the set of complex numbers}}

Step-by-step explanation:

18+128n^2=0\qquad\text{subtract 18 from both sides}\\\\128n^2=-18\qquad\text{divide both sides by 128}\\\\n^2=-\dfrac{18}{128}\\\\n^2=-\dfrac{18:2}{128:2}\\\\n^2=-\dfrac{9}{64}

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To get this, we use Pythagoras' theorem for the base

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According ro Pythagoras' ; the square of the hypotenuse equals the sum of the squares of the two other sides

Let us have the diagonal as l

Mathematically;

\begin{gathered} l^2=6^2+8^2 \\ l^2\text{ = 36 + 64} \\ l^2\text{ =100} \\ l\text{ = }\sqrt[]{100} \\ l\text{ = 10 mm} \end{gathered}

Now, to get the diagonal, we use the triangle with height 5 mm and the base being the hypotenuse we calculated above

Thus, we calculate this using the Pytthagoras' theorem as follows;

\begin{gathered} d^2=5^2+10^2 \\ d^2\text{ = 25 + 100} \\ d^2\text{ = 125} \\ d\text{ = }\sqrt[]{125} \\ d\text{ = }11.2\text{ mm} \end{gathered}

7 0
10 months ago
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3 0
2 years ago
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3 years ago
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Answer:

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