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Mathematics

1 answer:

NeX [460]3 years ago

7 0

Answer:

c = 50

Step-by-step explanation:

Using the Pythagorean Theorem:

Plug in the values of a = 14, and b = 48

$a^{2} +b^{2} =c^{2}\\14^{2} +48^{2} =c^{2} \\c^{2} = 2500\\c = \sqrt{2500} \\c = 50$

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Solve the equation 6w2 – 7w – 20 = 0.

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For this case we have the following quadratic equation:

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$w = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\w = \frac {- (- 7) \pm \sqrt {(- 7) ^ 2-4 (6) (- 20)}} {2 (6)}\\w = \frac {7 \pm \sqrt {49 + 480}} {12}\\w = \frac {7 \pm \sqrt {529}} {12}\\w = \frac {7 \pm23} {12}$

The roots are:

$w_ {1} = \frac {7 + 23} {12} = \frac {30} {12} = \frac {15} {6} = \frac {5} {2}\\w_ {2} = \frac {7-23} {12} = \frac {-16} {12} = - \frac {8} {6} = - \frac {4} {3}$