we know that
10*pi/3------> convert to degrees------> 10*180/3=600 degrees
600=360+240
240 degrees belong to the III quadrant
so
the x-coordinate is negative
the y-coordinate is negative
240°=180°+60°
60 degrees is the angle to find the terminal point
let
∅=60°
in the unit circle
r=1 units
x=-r*cos ∅----> x=-cos 60------> x=-(1/2)
y=-r*sin ∅----> y=-sin 60------> y=-(√3)/2
the point is
therefore
the answer is the option D
Answer:
(0.88, 0,65)
Step-by-step explanation:
what do we have to do ? I assume we need to find the crossing point between the 2 lines.
when looking at the graphic, we can already rule out the first, third and fourth option.
none of the coordinates of that point is larger than 1 in any direction or smaller than 0.5.
so, just by looking at things I immediately "bet" on answer option 2 (0.88, 0.65).
but let's check.
y = -2x/5 + 1
y = 3x - 2
for the crossing point both functions must produce the same functional value.
=>
-2x/5 + 1 = 3x - 2
-2x/5 + 5/5 = 15x/5 - 10/5
-2x + 5 = 15x - 10
15 = 17x
x = 15/17 = 0.88...
y = 3×0.88... -2 = 0.65...
hurray, our original observation and "suspicion" is confirmed.
Answer:
A) mJK = 21
B) mJNM = 159º
C) mKL = 159
D) mJKM = 201
E) mMKL = 339
Step-by-step explanation:
Circle = 360º
360º - 21º = 339º (what all missing angles have to add up to)
Assuming that:
JNL and KNM are both 180º angles
KJ and JM are supplementary as well as KL and LM
If we subtract 21º from 180º we get 159º which is both LM and KL
If we add 21º to 180º we get 201º which is both JKM and LMK
By subtracting 21º from 360º we know that MKL is 339ª
if we add 159º + 159º + 21º we get 339º which proves that those are our missing angles becuase our missing angles must make a total of 339º
Sorry that took so long I was pulling out all my notes from last year to make sure I was doing this right ^^;