See description in the pic.
Answer:
97km/h to 1dp
Step-by-step explanation:
47mins is 0.783 of an hour
76km/0.783 = 97.0212km/h
Answer:
The distance between the astronomers and the moon was ![7.56*10^{8]](https://tex.z-dn.net/?f=7.56%2A10%5E%7B8%5D)
meters.
Step-by-step explanation:
We have that the speed is the distance divided by the time, so:
In this problem, we have that:
The reflected laser beam was observed by the astronomers 2.52 s after the laser pulse was sent. This means that 
.
If the speed of light is 3.00 times 10^8 m/s, what was the distance between the astronomers and the moon?
We have that 
m/s.
We have to find d. So:
The distance between the astronomers and the moon was meters.
Answer:
answer=answer
Step-by-step explanation:
Y E S
I love simple math.
3.25$ to wash the car.
60 seconds: 1 quarter
how many quarters is 3.25?
4 quarters per dollar. so it'd be 13 quarters, then do 13*60=how many seconds u got. 780 seconds. I think you can do 780/60
HINT: hey, psst, it's 13 minutes. : D
HEY DOn'T teLL tHeM tHe ANSWER. :V
Answer:
Step-by-step explanation:
The diagram of the triangles are shown in the attached photo.
1) Looking at ∆AOL, to determine AL, we would apply the sine rule
a/SinA = b/SinB = c/SinC
21/Sin25 = AL/Sin 105
21Sin105 = ALSin25
21 × 0.9659 = 0.4226AL
AL = 20.2839/0.4226
AL = 50
Looking at ∆KAL,
AL/Sin55 = KL/Sin100
50/0.8192 = KL/0.9848
50 × 0.9848 = KL × 0.8192
KL = 49.24/0.8192
KL = 60
AK/Sin25 = AL/Sin 55
AKSin55 = ALSin25
AK × 0.8192 = 0.4226 × 50
AK = 21.13/0.8192
AK = 25.8
2) looking at ∆AOC,
Sin 18 = AD/AC = 18/AC
AC = 18/Sin18 = 18/0.3090
AC = 58.25
Sin 85 = AD/AB = 18/AB
AB = 18/Sin85 = 18/0.9962
AB = 18.1
To determine BC, we would apply Sine rule.
BC/Sin77 = 58.25/Sin85
BCSin85 = 58.25Sin77
BC = 58.25Sin77/Sin85
BC = 58.25 × 0.9744/0.9962
BC = 56.98
