Answer:
Its infinite solutions because its the same exact thing when you simplify the first half of the problem.
Step-by-step explanation:
Since the big triangle is an isosceles one, we conclude that the angles of the left triangle are: 32, 90 and 3x-10. Sinche the sum of this angles have to be 180° we have:
![(3x-10)+32+90=180](https://tex.z-dn.net/?f=%283x-10%29%2B32%2B90%3D180)
Solving for x, we have:
![\begin{gathered} (3x-10)+32+90=180 \\ 3x-10+32+90=180 \\ 3x=180+10-90-32 \\ 3x=68 \\ x=\frac{68}{3} \\ x=22.67 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%283x-10%29%2B32%2B90%3D180%20%5C%5C%203x-10%2B32%2B90%3D180%20%5C%5C%203x%3D180%2B10-90-32%20%5C%5C%203x%3D68%20%5C%5C%20x%3D%5Cfrac%7B68%7D%7B3%7D%20%5C%5C%20x%3D22.67%20%5Cend%7Bgathered%7D)
Therefore, x=22.67 and the answer is A.
Answer:
2
Step-by-step explanation:
h(4)=-1
g(h(4))=g(-1)=3
f(3)=2
We have been given a diagram. We are asked find the measure of arc EAB.
First of all, we will find the measure of arcs ED and CB using our given information.
We know that measure of an inscribed angle is half the measure of intercepted arc.
We can see that angle EBC is inscribed angle of arc EDC, so measure of arc EDC will be twice the measure of angle EBC.
![\widehat{EDC}=2\times m\angle EBC](https://tex.z-dn.net/?f=%5Cwidehat%7BEDC%7D%3D2%5Ctimes%20m%5Cangle%20EBC)
![\widehat{EDC}=2\times 80^{\circ}](https://tex.z-dn.net/?f=%5Cwidehat%7BEDC%7D%3D2%5Ctimes%2080%5E%7B%5Ccirc%7D)
![\widehat{EDC}=160^{\circ}](https://tex.z-dn.net/?f=%5Cwidehat%7BEDC%7D%3D160%5E%7B%5Ccirc%7D)
Similarly, we will find the measure of arc DCB.
![\widehat{DCB}=2\times m\angle DEB](https://tex.z-dn.net/?f=%5Cwidehat%7BDCB%7D%3D2%5Ctimes%20m%5Cangle%20DEB)
![\widehat{DCB}=2\times 70^{\circ}](https://tex.z-dn.net/?f=%5Cwidehat%7BDCB%7D%3D2%5Ctimes%2070%5E%7B%5Ccirc%7D)
![\widehat{DCB}=140^{\circ}](https://tex.z-dn.net/?f=%5Cwidehat%7BDCB%7D%3D140%5E%7B%5Ccirc%7D)
![\widehat{ED}=\widehat{EDC}-\wdiehat{DC}](https://tex.z-dn.net/?f=%5Cwidehat%7BED%7D%3D%5Cwidehat%7BEDC%7D-%5Cwdiehat%7BDC%7D)
![\widehat{ED}=160^{\circ}-88^{\circ}](https://tex.z-dn.net/?f=%5Cwidehat%7BED%7D%3D160%5E%7B%5Ccirc%7D-88%5E%7B%5Ccirc%7D)
![\widehat{ED}=72^{\circ}](https://tex.z-dn.net/?f=%5Cwidehat%7BED%7D%3D72%5E%7B%5Ccirc%7D)
![\widehat{ED}+\widehat{DCB}+\widehat{EAB}=360^{\circ}](https://tex.z-dn.net/?f=%5Cwidehat%7BED%7D%2B%5Cwidehat%7BDCB%7D%2B%5Cwidehat%7BEAB%7D%3D360%5E%7B%5Ccirc%7D)
![72^{\circ}+140^{\circ}+\widehat{EAB}=360^{\circ}](https://tex.z-dn.net/?f=72%5E%7B%5Ccirc%7D%2B140%5E%7B%5Ccirc%7D%2B%5Cwidehat%7BEAB%7D%3D360%5E%7B%5Ccirc%7D)
![212^{\circ}+\widehat{EAB}=360^{\circ}](https://tex.z-dn.net/?f=212%5E%7B%5Ccirc%7D%2B%5Cwidehat%7BEAB%7D%3D360%5E%7B%5Ccirc%7D)
![\widehat{EAB}=360^{\circ}-212^{\circ}](https://tex.z-dn.net/?f=%5Cwidehat%7BEAB%7D%3D360%5E%7B%5Ccirc%7D-212%5E%7B%5Ccirc%7D)
![\widehat{EAB}=148^{\circ}](https://tex.z-dn.net/?f=%5Cwidehat%7BEAB%7D%3D148%5E%7B%5Ccirc%7D)
Therefore, the measure of arc EAB is 148 degrees and option C is the correct choice.