If angle a is congruent to itself by the reflexive property, which transformation could be used to prove δabc ~ δade by aa similarity postulate?

triangles abc and ade, in which point d is between points a and b and point e is
between points a and c

a) translate triangle ade so that point d lies on point b to confirm ∠d ≅ ∠b.

b) translate triangle ade so that point b lies on point e to confirm ∠b ≅ ∠e.

c) dilate δabc from point a by the ratio segment ad over segment ab to confirm segment ad ~ segment ab.

d) dilate δabc from point a by the ratio segment ae over segment ac to confirm segment ae ~ segment ac.

5050

hope this !

0.0076, if you meant 4.125 divided by 0.03125 it would be 132

step-by-step explanation:

first you have to move the decimal in 0.03125 over until its behind the 5 which is 5 places over. now the number is 3125

then you have to move the decimal in 4.125, 5 places over. now the number is 4,112,500.

now you divide like you would normally would and get 0.0076

it should be a.

hope it

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.