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bagirrra123 [75]
3 years ago
6

Solve the equation 14 = 14n

Mathematics
2 answers:
igor_vitrenko [27]3 years ago
8 0

Answer: N=1

Step-by-step explanation:

Setler [38]3 years ago
7 0

Answer:

1=n

Step-by-step explanation:

Rewrite the equation as

14=14n

Divide each term by

14

and simplify.

Divide each term in

14

n=14

by 14

.14n

14=14/14

Cancel the common factor of

14

Cancel the common factor.

14

n14=

14/14

Divide

n by 1.n=14/14

Divide

14 by 14

.n=1

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Describe the solutions for the quadratic equation shown below:
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The solutions for the quadratic equation 9x² - 6x + 5 = 0 are A. 2 complex roots

To determine the the type of roots the quadratic equation 9x² - 6x + 5 = 0, we use the quadratic formula to find the roots.

So, for a quadratic equation ax + bx + c = 0, the roots are

x = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}

With a = 9, b = -6 and c = 5, the roots of our equation are

x = \frac{-(-6) +/- \sqrt{(-6)^{2} - 4 X 9 X 5} }{2 X 9} \\x = \frac{6 +/- \sqrt{36 - 180} }{18} \\x = \frac{6 +/- \sqrt{-144} }{18} \\x = \frac{6 +/- \sqrt{-12} }{18}\\x = \frac{6}{18} +/- i\frac{12}{18} \\x = \frac{1}{3} +/- i\frac{2}{3} \\x = \frac{1 + 2i}{3}  or \frac{1 - 2i}{3}

Since the roots of the equation are (1 + 2i)/3 and (1 - 2i)/3, there are 2 complex roots.

So, the solutions for the quadratic equation 9x² - 6x + 5 = 0 are A. 2 complex roots

Learn more about quadratic equations here:

brainly.com/question/18117039

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Are my answers for the conditional statement correct? I would just like to check them and if they're incorrect, could you let me
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We can see from the statement that p = brush teeth twice a day and q = not get cavities.

Thus:

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If this confuses you, try the example of p = cat and q = four legs. (i.e. the statement is If this is a cat, it has four legs). Try making the inverse, converse, and contrapositive, and see the truth of those statements.

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Give an example of a rational function that has a horizontal asymptote at y = 1 and a vertical asymptote at x = 4.
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Possibility 2: \frac{(x+1)(x+3)(x-3)}{x(x-4)(x+5)}

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There are infinitely many more possibilities.

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Possibility 3: \frac{-4(x-4)(x+1)}{-4(x-4)(x-4)}

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