a) C(0.5) = 39, C(2) = 39, C(4) = 69
b) Answers in part (a) represents the values of C at x = x₀
c) C(x) is continuous at x = 2
C (x) = 39 for 0 ⩽ x ⩽ 2
Since x = 0.5 is in the range 0 ⩽ x ⩽ 2:
C(0.5) = 39
Since x = 2 is in the range 0 ⩽ x ⩽ 2
C(2) = 39
Since x = 4 is in the range x > 2
C(4) = 39 + 15(x - 2)
C(4) = 39 + 15 (4 - 2)
C(4) = 39 + 15(2)
C(4) = 39 + 30
C(4) = 69
b) The answers in part A represents C(x). That is, the values of C at x = x₀
c) Is C(x) continuous at x = 2
C(2) = 39
Since 
, the function C(x) is continuous at x = 2
Learn more here: brainly.com/question/20710468
Answer:
C.)
Step-by-step explanation:
4/20=8/x
In an arithmetic equation, there is no variable in the 'meat' of the equation(example: 5-5=0). In an algebraic equation, there is a variable in the meat of the equation(example: 5-x=0).
Hope this helps and please give brainliest!
Draw a triangle with shortest side on the ground (horiz.). The length of this side is 80 feet. Two angles touch the ground: a right angle, and the 70 degree angle.
Then the height of the building is the 2nd shortest side.
The tangent function is the appropriate one to use here:
opp h
tan 70 deg = -------- = ----------
adj 80 ft
h
Since tan 70 deg = 2.75, 2.75 = -------
80 ft
so h = building height = (80 ft)(2.75) = 219.8 ft
