Answer:
Since the length of the drawing is 200 ft. and equivalent to 13.33 in. with a scale of 15 ft to 1 in. and the length of the paper is 11 in., Adoncia's drawing will not fit on the sheet of paper
Step-by-step explanation:
The given parameters are;
The scale of the drawing is 15 ft = 1 in.
The actual dimensions of the monument;
Height = 80 ft.
Length = 200 ft.
Therefore, we have;
The required dimension of the paper height = 80/15 = 16/3 = 5.33 in.
The required dimension of the paper length = 200/15 = 40/3 = 13.33 in.
The given paper dimension by 11 in. which is of a dimension of that of a standard letter paper size of 8.5 in. by 11 in.
Drawing length, 13.33 in. > Paper length > 11 in.
Adoncia's drawing will not fit on the sheet of paper.
Answer:
sin−1(StartFraction 8.9 Over 10.9 EndFraction) = x
Step-by-step explanation:
From the given triangle JKL;
Hypotenuse KJ = 10.9
Length LJ is the opposite = 8.9cm
The angle LKJ is the angle opposite to side KJ = x
Using the SOH CAH TOA Identity;
sin theta = opp/hyp
sin LKJ = LJ/KJ
Sinx = 8.9/10.9
x = arcsin(8.9/10.9)
sin−1(StartFraction 8.9 Over 10.9 EndFraction) = x
2,070mg 86%
— = —
? mg 100%
Cross multiply (2,070 X 100) which equals 207,000
Then divide 207,000 by 86 which equals
2,406.976
Then you round so you get 2,407
The Answer is 2,407 milligrams
Answer:
48
Step-by-step explanation:
120*(5/5-3/5)
120*(2/5)
24*2
48
