In ΔFGH, \overline{FH} FH is extended through point H to point I, \text{m}\angle FGH = (x-2)^{\circ}m∠FGH=(x−2) ∘ , \text{m}\angle GHI = (4x-5)^{\circ}m∠GHI=(4x−5) ∘ , and \text{m}\angle HFG = (x+19)^{\circ}m∠HFG=(x+19) ∘ . Find \text{m}\angle HFG. M∠HFG.

first one

step-by-step explanation:

answer: plz put a pic in so i can answer it

step-by-step explanation:

step-by-step explanation:

well if you round, things become much more simple. round 5.95 to 6, 4.25 to 4, and 1.05 to 1. now, add 6+4+1 and that would equal 11. so the items together would equal to about $11.

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.