What is the distance between the points (1.6,0.1) and (2.4,1.6)?

1.7 units

Step-by-step explanation:

Let (1.6,0.1) be (x1,y1) and (2.4,1.6) be (x2,y2)

By using distance formula,

d= sqrt (x2-x1)^2 + (y2-y1)^2

=sqrt (2.4-1.6)^2 + (1.6-0.1)^2

=sqrt (0.8)^2 + (1.5)^2

=sqrt 0.64+2.25

=sqrt 2.89

=sqrt (1.7)^2

=1.7

Therefore,the distance between them is 1.7 units

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