PLEASE HELP!! Mark the givens in the diagram. Then copy the proof on paper, and fill in the blanks to complete the proof. If angle S is congruent to angle I. angle G is congruent to angle A and T is the midpoint of SI prove SG is congruent to IA

S is congruent to
ST is congruent to
Angle G is congruent to Angle
STG is congruent to
SG is congruent to

hello from mrbilldoesmath!

answer:   x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb

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.