Between 2005 -2009, the percent (in decimal form)of network commercials that lasted exactly 18 seconds could be modeled by the equation y=0.38x2-1.8\0.56x2+100,where x is the number of years since 2005 what percent of network commercials lasted exactly 18 seconds in 2008?Round you answer to the nearest percent

Approximately 1.54% of the commercials lasted exactly 18 seconds in 2008.

Step-by-step explanation:

We are given the following in the question:

y is the percent of network commercials that lasted exactly 18 seconds and x is the number of years since 2005.

We have to estimate the percentage of advertisement that  lasted exactly 18 seconds in 2008.

Thus, we put x = 3 in the equation:

Thus, approximately 1.54% of the commercials lasted exactly 18 seconds in 2008.

that is the sum

step-by-step explanation:

there are no other answers without knowing the variables values

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.