Statement reason 1. draw a line through b that is perpendicular to and label the point of intersection with as d. construction 2. in δabd, bd = c sin a. definition of sine in a right triangle 3. in δcbd, bd = a sin c. definition of sine in a right triangle 4. c sin a = a sin c substitution property of equality 5. dividing throughout by sin a sin c 6. draw a line through a that is perpendicular to and label the point of intersection with as e. construction 7. in δbae, ae = c sin b. definition of sine in a right triangle 8. in δcae, ae = b sin c. definition of sine in a right triangle 9. c sin b = b sin c substitution property of equality 10. dividing throughout by sin b sin c 11. transitive property of equality what is the missing statement in step 10 of the proof? given: δabc with ab = c, bc = a, ac = b prove:

v v cj van config k tbs eng r.c r.c fcb fb gb frfr

80° is the measure of gh

step-by-step explanation:

circle f is congruent to circle j

we have two circle with center f and j.

both circles are congruent. the radius of the circle would be equal.

see the attachment for the circle.

we are given ∠efd ≅ ∠gjh

therefore, arc (ed) = arc (gh)     (∴ circle f is congruent to circle j)

but ∠dfe=80°=arc (ed)

therefore, arc (gh) = 80°

thus, 80° is the measure of gh

step-by-step explanation:

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.