Make a circle with the following measurements:
1 answer:
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Answer:
The correct options are;
EFGH has 4 congruent sides
Diagonal FH bisects angles EFG and EHG
Angle FEH is congruent to angle FGH
Step-by-step explanation:
1) Given that for a reflection, we have;
The distance of the reflected preimage from the line of reflection = The distance of the reflected image from the line of reflection
Therefore;
The distance of the point E from the line HF = The distance of the point G from the line HF
Also the reflection of an preimage (x, y) about the x-axis, gives an image (x, -y)
We can show that from the length of a line given by the equation, that the length EH ≅ GH and EF ≅ GF
Therefore since we are given that EH = EF, we have;
EH = GH = GF = EF by the definition of congruency, which gives 4 congruent sides
2) Given that EH = GH = GF = EF and HF = FH by reflective property, we have;
ΔEHF ≅ ΔGHF
∴ ∠GHF ≅ ∠EHF by Congruent Parts of Congruent Triangles are Congruent
Similarly, ∠GFH ≅ ∠EFH
Therefore, ∠GFH = ∠EFH and ∠GHF = ∠EHF
Therefore, diagonal FH bisects angles EFG and EHG
3) Given that ΔEHF ≅ ΔGHF, we have;
Angle FEH is congruent to angle FGH, by Congruent Parts of Congruent Triangles are Congruent
All we need is to put this form in the vertex form f(x) = (ax+b)^2 + c
So we have <span>f (x)= 3x^2+12x+11 ....
Let's complete the square (if you aware of it)
</span><span>
f(x)= 3x^2+12x+11 = 3(x^2+4x)+11 = 3(x^2+4x+4-4)+11
=</span><span> 3([x^2+4x+4]-4)+11 = 3[(x+2)^2-4]+11 =3</span><span>(x+2)^2 - 12 +11 = 3</span><span><span>(x+2)^2 -1
so our form would be:
Here is a parabola with vertex of (-2,-1) and with positive </span> slope (concave up)
</span>
So we have <span>f (x)= 3x^2+12x+11 ....
Let's complete the square (if you aware of it)
</span><span>
f(x)= 3x^2+12x+11 = 3(x^2+4x)+11 = 3(x^2+4x+4-4)+11
=</span><span> 3([x^2+4x+4]-4)+11 = 3[(x+2)^2-4]+11 =3</span><span>(x+2)^2 - 12 +11 = 3</span><span><span>(x+2)^2 -1
so our form would be:
Here is a parabola with vertex of (-2,-1) and with positive </span> slope (concave up)
</span>
I hope that helps!