3 years ago

Can someone please help me with this question? Thank you!

Mathematics

1 answer:

Goryan [66]3 years ago

4 0

Answer: A

<u>Step-by-step explanation:</u>

$\stackrel\frown{XY}$ ≅ $\stackrel\frown{YZ}$

⇒ ∠XOY ≅ ∠ZOY

$\overline{XO}$ ≅ $\overline{ZO}$ since they are both a radius

$\overline{OY}$ ≅ $\overline{OY}$ based on reflexive property

∴ ΔXOY ≅ ΔZOY by SAS

So, $\overline{XY}$ ≅ $\overline{ZY}$ by CPCTC

ZY = 10.2 is given

so, XY = 10.2 by substitution property

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Solve 2x^2 + x - 4 = 0 <br> X2 +

damaskus [11]

Answer:

$\large \boxed{\sf \ \ x = -\dfrac{\sqrt{33}+1}{4} \ \ or \ \ x = \dfrac{\sqrt{33}-1}{4} \ \ }$

Step-by-step explanation:

Hello, please find below my work.

$2x^2+x-4=0\\\\\text{*** divide by 2 both sides ***}\\\\x^2+\dfrac{1}{2}x-2=0\\\\\text{*** complete the square ***}\\\\x^2+\dfrac{1}{2}x-2=(x+\dfrac{1}{4})^2-\dfrac{1^2}{4^2}-2=0\\\\\text{*** simplify ***}\\\\(x+\dfrac{1}{4})^2-\dfrac{1+16*2}{16}=(x+\dfrac{1}{4})^2-\dfrac{33}{16}=0$

$\text{*** add } \dfrac{33}{16} \text{ to both sides ***}\\\\(x+\dfrac{1}{4})^2=\dfrac{33}{16}\\\\\text{**** take the root ***}\\\\x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4}\\\\\text{*** subtract } \dfrac{1}{4} \text{ from both sides ***}\\\\x = -\dfrac{1}{4} -\dfrac{\sqrt{33}}{4} \ \ or \ \ x = -\dfrac{1}{4} +\dfrac{\sqrt{33}}{4}$