+63 can someone tell me what I am doing wrong
(sorry for the bad picture I tried get one the best i could.)
+701 not really sure what this is but,,, a tip i would give is to simplify the equations to get plain x so idrk...
maybe that's why you got it wrong, but i don't know...just simplify for example the 14x=64
please correct me if i am wrong, and i am sorry i wasn't much help :') but i hope this fixes your problem :"))
+63 The topic is inscribed angles. I have to find the either the arcs or the angles of a circle. As far as I know none of the problems are supposed to be decimal answers.
+701 ok so basically where you went wrong is when you multiplied by two instead of dividing by two...
+63 well yes and no. Like I did \(16x-10=1/2(67)\)
I still got a decimal answer. What did I do wrong?
Wait i figured it out. Thanks for the help.
The basic theorem that's used here is,
' the angle at the centre of the circle is twice the angle at the circumference '.
So, in Q9, the angle at the centre TWV, will be twice the angle at the circumference TUV,
ie. angle TWV = 2 times 67 = 134.
It follows then, that 16x - 10 = 134,
16x = 144,
x = 144/16 = 9.
In Q10, angle KNM = 2 times (7x + 9) etc.
Q's 11 and 12 require a little more thought. Come back if you can't do them, (after you've tried to do them that is).
Basically, the situation here to be understood is that an angle inscribed (vertex on the circle perimeter) intercepts an arc ofTWICE the degrees of the angle .
#9 16x-10 = 2 (67) solve for x then put x back into the 16x-10 to find the arc degrees
#10 2 (7x+9) = 46
#11 2 (15x+2) = 360 - 87-39
#12 360 - ( 75 + 2 (59)) = 17x-20
Cool? 
+37158 Basically, the situation here to be understood is that an angle inscribed (vertex on the circle perimeter) intercepts an arc ofTWICE the degrees of the angle .
#9 16x-10 = 2 (67) solve for x then put x back into the 16x-10 to find the arc degrees
#10 2 (7x+9) = 46
#11 2 (15x+2) = 360 - 87-39
#12 360 - ( 75 + 2 (59)) = 17x-20
Cool?
