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3 years ago

Congruent Triangles and Proofs Given: JK bisects

Mathematics

1 answer:

Nikitich [7]3 years ago

7 0

Answer:

Step-by-step explanation:

1S. JM bisects ∠ KJL and ∠JMK ≅ ∠JML

1R. Given

2S. ∠KJM ≅ ∠LJM

2R. Angle bisectors form ≅ ∠s

3S. JM≅JM

3R. Reflexive Propriety ( segment is ≅ with itself)

4S. ΔJMK ≅ΔJML

4R. ASA theorem of congruency

5S. JK≅JL

5R. Coresponding parts of congruent Δs are congruent (CPCTC)

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lorasvet [3.4K]

Answer:

$x = \sqrt{2} - 8\\x = -\sqrt{2} - 8$

Step-by-step explanation:

To complete the square, we first have to get our equation into $ax^2 + bx = c$ form.

First we add 16x to both sides:

$x^2 + 16x + 62 = 0$

And now we subtract 62 from both sides.

$x^2 + 16x = -62$

We now have to add $(\frac{b}{2})^2$ to both sides of the equation. b is 16, so this value becomes $(16\div2)^2 = 8^2 = 64$ .

$x^2 + 16x + 64 = -62+64$

We can now write the left side of the equation as a perfect square. We know that x+8 will be the solution because $8\cdot8=64$ and $8+8=16$ .

$(x+8)^2 = -62 + 64$

We can now take the square root of both sides.

$x+8 = \sqrt{-62+64}\\\\ x+8 = \pm \sqrt{2}$

We can now isolate x on one side by subtracting 8 from both sides.

$x = \pm\sqrt{2} - 8$

So our solutions are

$x = \sqrt{2} - 8\\x = -\sqrt{2} - 8$

Hope this helped!

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Step-by-step explanation:

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