Answer:
y = 4/3x - 10
Step-by-step explanation:
Slope: 4/3
y-intercept: -2 - (4/3)(6) = -2 - 8 = -10
Answer:
(xM, yM) (0, 0)
Step-by-step explanation:
Im not sure its right,
Answer:
A
Step-by-step explanation:
9/2 as a mixed fraction is
4 1/2
Hope this answers your question!
Solution:
(a) In the given figure,
![\angle ABO=x](https://tex.z-dn.net/?f=%5Cangle%20ABO%3Dx)
BD is the bisector of angle ABC, thus:
![\angle ABD=\angle DBC](https://tex.z-dn.net/?f=%5Cangle%20ABD%3D%5Cangle%20DBC)
So, write as follows:
(i)
![\angle ABC=\angle ABD+\angle DBC](https://tex.z-dn.net/?f=%5Cangle%20ABC%3D%5Cangle%20ABD%2B%5Cangle%20DBC)
![\begin{gathered} \angle ABC=\angle ABD+\angle ABD \\ \angle ABC=2\angle ABD \\ \angle ABC=2\angle ABO \\ \angle ABC=2x \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cangle%20ABC%3D%5Cangle%20ABD%2B%5Cangle%20ABD%20%5C%5C%20%5Cangle%20ABC%3D2%5Cangle%20ABD%20%5C%5C%20%5Cangle%20ABC%3D2%5Cangle%20ABO%20%5C%5C%20%5Cangle%20ABC%3D2x%20%5Cend%7Bgathered%7D)
Therefore,
![\operatorname{\angle}ABC=2x](https://tex.z-dn.net/?f=%5Coperatorname%7B%5Cangle%7DABC%3D2x)
(ii)
OB=OA=OD radii of the same circle.
As OB=OA then in triangle AOB,
![\angle ABO=\angle BAO=x](https://tex.z-dn.net/?f=%5Cangle%20ABO%3D%5Cangle%20BAO%3Dx)
BOD is a diameter of a circle. then by theorem of circle,
![\angle BAD=90^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20BAD%3D90%5E%7B%5Ccirc%7D)
thus,
![\angle OAD=\angle BAD-\angle OAB](https://tex.z-dn.net/?f=%5Cangle%20OAD%3D%5Cangle%20BAD-%5Cangle%20OAB)
![\angle OAD=90^{\circ}-x](https://tex.z-dn.net/?f=%5Cangle%20OAD%3D90%5E%7B%5Ccirc%7D-x)
OA=OD , in the tringle AOD,
![\angle OAD=\angle ODA=90^{\circ}-x](https://tex.z-dn.net/?f=%5Cangle%20OAD%3D%5Cangle%20ODA%3D90%5E%7B%5Ccirc%7D-x)
In triangle AOD,
![\angle AOD+\angle OAD+\angle ODA=180^{\circ}](https://tex.z-dn.net/?f=%5Cangle%20AOD%2B%5Cangle%20OAD%2B%5Cangle%20ODA%3D180%5E%7B%5Ccirc%7D)
![\begin{gathered} \angle AOD+90^{\circ}-x+90^{\circ}-x=180^{\circ} \\ \angle AOD+180^{\circ}-2x=180^{\circ} \\ \angle AOD=2x \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cangle%20AOD%2B90%5E%7B%5Ccirc%7D-x%2B90%5E%7B%5Ccirc%7D-x%3D180%5E%7B%5Ccirc%7D%20%5C%5C%20%5Cangle%20AOD%2B180%5E%7B%5Ccirc%7D-2x%3D180%5E%7B%5Ccirc%7D%20%5C%5C%20%5Cangle%20AOD%3D2x%20%5Cend%7Bgathered%7D)
Therefore,
![\begin{equation*} \angle AOD=2x \end{equation*}](https://tex.z-dn.net/?f=%5Cbegin%7Bequation%2A%7D%20%5Cangle%20AOD%3D2x%20%5Cend%7Bequation%2A%7D)
(iii) By the theorem of circle, the angle formed at the center of the circle is twice the angle formed at the circumference of the circle with the same base.
![\angle AOC=2\angle ABC](https://tex.z-dn.net/?f=%5Cangle%20AOC%3D2%5Cangle%20ABC)
![\begin{gathered} \angle AOC=2\times2x \\ \angle AOC=4x \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cangle%20AOC%3D2%5Ctimes2x%20%5C%5C%20%5Cangle%20AOC%3D4x%20%5Cend%7Bgathered%7D)
Therefore,
![\operatorname{\angle}AOC=4x](https://tex.z-dn.net/?f=%5Coperatorname%7B%5Cangle%7DAOC%3D4x)
(iv)
OA=OD , thus tringle AOD is an isosceles triangle,
![\angle OAD=\angle ODA=90^{\circ}-x](https://tex.z-dn.net/?f=%5Cangle%20OAD%3D%5Cangle%20ODA%3D90%5E%7B%5Ccirc%7D-x)
![\begin{gathered} \angle ADO=90^{\circ}-x \\ \angle ADB=90^{\circ}-x \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cangle%20ADO%3D90%5E%7B%5Ccirc%7D-x%20%5C%5C%20%5Cangle%20ADB%3D90%5E%7B%5Ccirc%7D-x%20%5Cend%7Bgathered%7D)
Therefore,