To answer this question we will have to make use of <u>the Isosceles Triangle Theorem</u> which states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle. Thus, as a corollary we know that is EF is the angle bisector of the vertex angle ∠E, then, EF is the perpendicular bisector of the of the base DK.

Please follow the diagram of a complete understanding of the logic and the solution.

As EF is the angle bisector as given in the question, thus we will have:

$m\angle DEK=2\times m\angle DE F=2\times 43^{\circ}=86^{\circ}$ .

Also, from the Theorem we know that KF will be half of DK and thus, KF will be: $KF=\frac{1}{2}DK= \frac{1}{2}\times 16=8$ centimeters.

Likewise, from the same theorem we have: $m\angle EFD=90^{\circ}$

4 0

2 years ago

Y=e^x how do I solve this

kolezko [41]

If you're trying to solve for x, keep in mind the ln (natural log) of e is 1. So take the natural log of both sides, which will give you ln Y = ln e^x. The ln and e essentially cancel out, so you're left with x = ln Y.

6 0

3 years ago