<span>Let A be the center of a circle and two angles at the adjacent center AOB and BOC. Knowing the measure of the angle AOB = 120 and the measure BOC = 150, find the measures of the angles of the ABC triangle.
</span>solution
Given the above information;
AC=AB, therefore ABC is an isosceles triangle.
therefore, BAO=ABO=(180-120)/2=30
OAC=OCA=(180-90)/2=45
OBC=BCO=(180-150)/2=15
THUS;
BAC=BAO+OAC=45+30=75
ABC=OBA+CBO=15+30=45
ACB=ACO+BCO=15+45=60
And?? More information to answer this..
Answer:
Angle f= 66
DE=10
Step-by-step explanation:
So first I think it's an isosceles triangle to 2 sides will have the same length.
For the angle I added 90 and 24 and subtracted them from 180.
Hello : let : y = f(x) so : y' = f'(x)
calculate : f'(x) and : f'(-1) given f(-1) = 2
by derivate : <span>4x3+2y2−11=4xy−x
12x² +4yy' = 4y +4xy' - 1
4yy' - 4xy' = - 12x² +4y -1
y' ( 4y - 4x) = 4y -12x² -1 ...(*)
if x = -1 y = 2 subsct in : (*)
y' (4(2)-4(-1)) = 4(2)-12(-1)² - 1
2y' = - 9
y' = - 9/2 = f'(- 1) ( the slope of the tangent )
</span>of the tangent line to the curve <span>at the point (−1,2) is :
y - 2 = (-9/2)(x +1)</span>
5-2 5/16
Convert the second term to improper fraction;
5-
=
=
=2
