A hhhhhhhhhhnjjjjnnnnnjjkkk
Answer:
Part a) The radii are segments AC and AD and the tangents are the segments CE and DE
Part b) 
Step-by-step explanation:
Part a)
we know that
A <u>radius</u> is a line from any point on the circumference to the center of the circle
A <u>tangent</u> to a circle is a straight line which touches the circle at only one point. The tangent to a circle is perpendicular to the radius at the point of tangency.
In this problem
The radii are the segments AC and AD
The tangents are the segments CE and DE
Part b)
we know that
radius AC is perpendicular to the tangent CE
radius AD is perpendicular to the tangent DE
CE=DE
Triangle ACE is congruent with triangle ADE
Applying the Pythagoras Theorem

substitute the values and solve for CE





remember that
CE=DE
so

Answer:
I believe the answer to this question is: Slope-Intercept-Form;
Y= - 7/2x - 27/2.
A = h[(a+b)/2]
h = 2 in
a = 12 in
b = 13 in
Plugging in all the values...
A = 2 [(12+13)/2]
A = 2 [25/2]
cancel out those 2 Twos
A = 25in²
Answer:
B
Step-by-step explanation:
The sign of the leading coefficient in both is positive so they both open up