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Answer:
A) ∠DHG = 96°
B) arc EF = 134°
Step-by-step explanation:
<u>Part A</u>:
As you have shown in your work, angle DHG is supplementary to given angle FHG:
∠DHG = 180° -∠FHG
∠DHG = 180° -84°
∠DHG = 96°
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<u>Part B</u>:
∠DHG is half the sum of arcs EF and DG.
96° = (1/2)(arc EF + 58°) . . . . . fill in known values
192° -58° = arc EF . . . . . . . . . . multiply by 2, subtract 58°
arc EF = 134°
Answer:
a) H0 = 15 m, is the initial height, the height of the rocket at time 0 s.
b) ![D=[0,5]](https://tex.z-dn.net/?f=D%3D%5B0%2C5%5D)
c) ![R=[0,35]](https://tex.z-dn.net/?f=R%3D%5B0%2C35%5D)
Step-by-step explanation:
a) H0 is the initial height. It means the height of the rocket at time 0 s.
This value is 15 m.
b) The domain is all values of t between 0 and 5 seconds.
c) The range is all values of height from 0 to 35 meters.
I hope it helps you!
9514 1404 393
Answer:
1/12 box
Step-by-step explanation:
Each brother gets 1/4 of the remaining 1/3 box, so ...
(1/4)(1/3 box) = 1/(4·3) box = 1/12 box
Each brother gets 1/12 of the original box.
Answer:
Last one
Step-by-step explanation:
In order to have a perpendicular bisector, everything needs to be even and connected.
Really hope this helps!
Sometimes
LCM of 3 and 4 is 12....(3 x 4)
LCM of 10 and 6 is 30.....not (10 x 6) = 60
