Given: ΔWXY is isosceles with legs WX and WY; ΔWVZ is isosceles with legs WV and WZ. Prove: ΔWXY ~ ΔWVZ
2 answers:
Answer:
By AA
ΔWXY ~ΔWVZ
Step-by-step explanation:
Here WXY is an isosceles triangle with legs WX & WY
So WX = WY
Hence ∠X = ∠Y
So ∠2= ∠3.
Now by angle sum property
∠1 + ∠2+∠3 = 180°
∠1+∠2+∠2=180°
2∠2 = 180° - ∠1 .......(1)
In triangle WVZ
WV = WZ
So ∠V = ∠Z
∠4 = ∠5
Once again by angle sum property
∠1 + ∠4 + ∠5=180°
∠1 + ∠4 + ∠4 = 180°
2∠4 = 180° - ∠1 ...(2)
From (1) & (2)
2∠2 = 2∠4
∠2=∠4
Now ∠W is common to both triangles
Hence by AA
ΔWXY ~ΔWVZ
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Answer:
Step-by-step explanation:
Begin with substuting the x variable with -2, we do this because the question has listed the value of x already.
Using the value of x, -2 we determine g(x).
g(x) = -2^2 + 2
Above is what the equation would look as, after you input the value of -2.
Using pemdas, (parantheses, exponents, multiplication, division, addition, subtraction) solve the equation.
-2^2 = 4
Think of it as -2 * -2, which is why -2^2 is 4.
Add 4 +2.
4 + 2 = 6.
Therefore, the value of g(x) = 6