A townhouse in San Francisco was purchased for $80,000 in 1975. The appreciation of the building is modeled by the equation: A=80000(1.12)^t, where t represents time in years.

In what year was the building worth double it’s value in 1975?

Year:

In 1981 was the building worth double it’s value.

Step-by-step explanation:

Given : A townhouse in San Francisco was purchased for $80,000 in 1975. The appreciation of the building is modeled by the equation : , where t represents time in years.

To find : In what year was the building worth double it’s value in 1975?

Solution :

The amount is $80,000.

The building worth double it’s value in 1975.

i.e. amount became A=2(80000).

Substitute in the model,

Taking log both side,

i.e. Approx in 6 years.

So, 1975+6=1981

Therefore, in 1981 was the building worth double it’s value.

answer:   28.56 feet (approx)

step-by-step explanation:

since, triangle abc has measure of angle c equal to 55 degrees, measure of angle abc equal to 90 degrees, and length of bc equal to 20 feet.

we have to find out the measurement of segment ab.

therefore, tan 55° =

but bc = 20 feet.

⇒ tan 55 ° =  

⇒ ab = 20 × tan 55 °

⇒ ab = 20 × 1.42814800674

⇒ ab = 28.5629601348   ≈ 28.56 feet