Yes but there is no photo
Answer:
C
Step-by-step explanation:
Yes <em>because</em><em> </em><em>there</em><em> </em><em>are</em><em> </em><em>two</em><em> </em><em>pairs</em><em> </em><em>of</em><em> </em><em>congruent</em><em> </em><em>corresponding</em><em> </em><em>angles</em><em>.</em>
Answer:
74
Step-by-step explanation:
Say that arc JL going through M is arc E and JL going the other way is arc D
For the angle formed by two tangents, K=(1/2)(E-D)
64=E-D
Furthermore, angle K and central angle JCL (facing toward K) are supplementary, so 180-K=JCL=180-32=148
Thus, as the angles around angle C add up to 360, angle JCL (facing toward M) is 360-148=32+180=212
E is then 212
64=212-D
212-64=D=148
Thus, as JML is an inscribed angle, M=1/2(D)=1/2(148)=74
Answer:
Use the Pythagorean theorem for right triangles to solve this
a^2 + b^2 = c^2 where 'a' and 'b' are the legs of the triangle and 'c' is the hypotenuse
so:
8^2 + b^2 = 17^2 solve for the second leg, b
64 + b^2 = 289
b^2 = 289 - 64= 225
b = 15
