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Svetradugi [14.3K]
3 years ago
10

Factor 391, 291, 387, 113, 451, 37, 411, 239, and 459 to prime factors

Mathematics
1 answer:
Zigmanuir [339]3 years ago
3 0

Answer:

391 = 17 × 23

291 =  3 × 97

387 =  3 × 3 × 43  = 3^2 × 43

113 is prime

451 = 11 × 41

37 is prime

411 = 3 × 137

239 is prime

459 = 3 × 3 × 3 × 17  = =  3^3 × 17

Step-by-step explanation:

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Answer:

65

Step-by-step explanation:

https://www.gktoday.in/aptitude/the-next-number-of-the-sequence-3-5-9-17-33-is/


8 0
3 years ago
Identify the zeros of the function f(x) = 2x^2 − 4x + 5 using the Quadratic Formula
Nostrana [21]

For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.

Then, we must find the roots of:

2x ^ 2-4x + 5 = 0

Where:

a = 2\\b = -4\\c = 5

We have to:x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}

Substituting we have:

x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (5)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-40}} {4}\\x = \frac {4 \pm \sqrt {-24}} {4}

By definition we have to:

i ^ 2 = -1

So:

x = \frac {4 \pm \sqrt {24i ^ 2}} {4}\\x = \frac {4 \pm i \sqrt {24}} {4}\\x = \frac {4 \pm i \sqrt {2 ^ 2 * 6}} {4}\\x = \frac {4 \pm 2i \sqrt {6}} {4}\\x = \frac {2 \pm i \sqrt {6}} {2}

Thus, we have two roots:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

Answer:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

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Please help ! Easy 7th grade work .
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Answer:

1-9 answers

Step-by-step explanation:

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