Answer:
4k+ 23
Step-by-step explanation:
Answer:
1050 g < weight ≤ 1150 g
Step-by-step explanation:
Let w represent the weight of the package in grams. The the number of 100-gram increments after the first 250 grams is given by ...
⌈(w-250)/100⌉ . . . . . . . where ⌈ ⌉ signifies the <em>ceiling</em> function
and the charges for a package exceeding 250 grams will be ...
0.65 + 0.10⌈(w -250)/100⌉ = 1.55
0.10⌈(w -250)/100⌉ = 0.90 . . . . . . . . subtract 0.65
⌈(w -250)/100⌉ = 9 . . . . . . . . . . . . . . . divide by 0.10
8 < (w-250)/100 ≤ 9 . . . . . . . . . . . . . . meaning of ceiling function
800 < w -250 ≤ 900 . . . . . . . . . . . . . multiply by 100
1050 < w ≤ 1150 . . . . . . . . . . . . . . . . . add 250
The weight in grams could be greater than 1050 and at most 1150 for a charge of $1.55.
We can use law of cosines to find the length of the third side.
Law of cosines is:
c^2 = a^2 + b^2 - 2ab*cos(C), where angle C is opposite of side c.
Plug in what we know.
c^2 = 3^2 + 4^2 - 2(3)(4)cos(60)
Simplify:
c^2 = 9 + 16 - 24cos(60)
c^2 = 25 - 24cos(60)
Solve for c by taking the square root of both sides:
c = sqrt(25 - 24cos(60))
c = sqrt(25 - 24(0.5))
c = sqrt(13)
c = 3.606