An alternating current i in amperes in a certain circuit can be found after t seconds using i= 3 (sin 150)t , where 150 is a degree measure. Rewrite the formula in terms of sum of two angle measures. Group of answer choices

a 3[sin (130 + 15)] t

b 3[sin (45 + 45)] t

c 3[sin (180 - 45)] t

d 3[sin (90 + 60)] t

b

step-by-step explanation:

second option

step-by-step explanation:

the measures of a triangle equal 180. use that to find x.

2x+3x+4x=180

9x=180

9x/9=180/9

x=20

now you know x, you know angle a is 40°, angle b is 60°, angle c is 80°.

because angle a is 40°, the first option doesn't work.

because angle b is 60°, the second option does work.

when you add angle a and b, you get 100°. complementary means it equals 90°, so the third option doesn't work.

when you add angle a and c, you get 120°, which is not 100.

(2,6)

step-by-step explanation:

cause i just did this and got the answer

you're welcome

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.