The percent p of the u. s. population with internet access is given by 25p-34t=1378 where t is the number of years after 1990. what percent of the u. s. population had interent access in 2005? (round to one decimal place) in what year will 89% have internent access?

Step-by-step explanation:

the percent p of the u.s. population with internet access is given by:

25 p-34 t=1378

where t is the number of years after 1990.

⇒  25p-34×(15)=1378

⇒  25p-510=1378

⇒  25p=1378+510

⇒  25p=1888

⇒  p=75.52

Hence, 75.52 percent of the u.s. population had internet access in 2005.

i.e. we have to find the value of t when we are given p=89

⇒  25×89-34t=1378

⇒  2225-34t=1378

⇒  2225-1378=34t

⇒  847=34t

⇒  or 34t=847

⇒  t=24.9112≈25

Hence after 25 years 89% will have internet access.

i.e. in the year 2015 89% will have internet access.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

jeff jeff

step-by-step explanation: