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Alborosie
2 years ago
5

Experimentally verify sum of three angles of an triangle is 180⁰.​

Mathematics
1 answer:
svetlana [45]2 years ago
4 0

Draw line a through points A and B. Draw line b through point C and parallel to line a.

Since lines a and b are parallel, <)BAC = <)B'CA and <)ABC = <)BCA'.

It is obvious that <)B'CA + <)ACB + <)BCA' = 180 degrees.

Thus <)ABC + <)BCA + <)CAB = 180 degrees.

Lemma

If ABCD is a quadrilateral and <)CAB = <)DCA then AB and DC are parallel.

Proof

Assume to the contrary that AB and DC are not parallel.

Draw a line trough A and B and draw a line trough D and C.

These lines are not parallel so they cross at one point. Call this point E.

Notice that <)AEC is greater than 0.

Since <)CAB = <)DCA, <)CAE + <)ACE = 180 degrees.

Hence <)AEC + <)CAE + <)ACE is greater than 180 degrees.

Contradiction. This completes the proof.

Definition

Two Triangles ABC and A'B'C' are congruent if and only if

|AB| = |A'B'|, |AC| = |A'C'|, |BC| = |B'C'| and,

<)ABC = <)A'B'C', <)BCA = <)B'C'A', <)CAB = <)C'A'B'.

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5x3 + 14x2 + 9x <br> help
Cerrena [4.2K]
<h3>Answer:    x(x+1)(5x+9) </h3>

===================================================

Work Shown:

5x^3 + 14x^2 + 9x

x( 5x^2 + 14x + 9 )

To factor 5x^2 + 14x + 9, we could use the AC method and guess and check our way to getting the correct result.

A better way in my opinion is to solve 5x^2 + 14x + 9 = 0 through the quadratic formula

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(14)\pm\sqrt{(14)^2-4(5)(9)}}{2(5)}\\\\x = \frac{-14\pm\sqrt{16}}{10}\\\\x = \frac{-14\pm4}{10}\\\\x = \frac{-14+4}{10} \ \text{ or } \ x = \frac{-14-4}{10}\\\\x = \frac{-10}{10} \ \text{ or } \ x = \frac{-18}{10}\\\\x = -1 \ \text{ or } \ x = \frac{-9}{5}\\\\

Then use those two solutions to find the factorization

x = -1  or  x = -9/5

x+1 = 0  or  5x = -9

x+1 = 0  or  5x+9 = 0

(x+1)(5x+9) = 0

So we have shown that 5x^2 + 14x + 9 factors to (x+1)(5x+9)

-----------

Overall,

5x^3 + 14x^2 + 9x

factors to

x(x+1)(5x+9)

6 0
3 years ago
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