<span><span> Let ABC be an isosceles triangle with AB = AC. Let M denote the midpoint of BC (i.e., M is the point on BC for which MB = MC). Thena)<span> </span>Triangle ABM is congruent to triangle ACM.b)<span> </span>Angle ABC = Angle ACB (base angles are equal)c)<span> </span>Angle AMB = Angle AMC = right angle.d)<span> </span>Angle BAM = angle CAM
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Formula
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Find Area
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Answer: Area = 88m²-----------------------------
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)
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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)
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I hope that
helps!
its 25 because yeah ......e.w.cds.vfc.fd.cdvfd dvfvfd