Part 1
Part 2
Part 3
Part 4
Part 5
![[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\ \\ =(1-5)\cdot4=-4\cdot4=-16](https://tex.z-dn.net/?f=%5B%282-3%29%5E2-5%5D%5Ccdot4%3D%5B%28-1%29%5E2-5%5D%5Ccdot4%20%5C%5C%20%20%5C%5C%20%3D%281-5%29%5Ccdot4%3D-4%5Ccdot4%3D-16)
Part 6
Part 7
Part 8
Part 9
![(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\ \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\ \\ =36+4=40](https://tex.z-dn.net/?f=%288-2%29%5E2%2B%5Cfrac%7B1%7D%7B4%7D%5B4-3%286-10%29%5D%3D6%5E2%2B%5Cfrac%7B1%7D%7B4%7D%5B4-3%28-4%29%5D%20%5C%5C%20%20%5C%5C%20%3D36%2B%5Cfrac%7B1%7D%7B4%7D%5B4-%28-12%29%5D%3D36%2B%5Cfrac%7B1%7D%7B4%7D%284%2B12%29%3D36%2B%5Cfrac%7B1%7D%7B4%7D%2816%29%20%5C%5C%20%20%5C%5C%20%3D36%2B4%3D40)
Part 10
Part 11
![3-2[2^3+(-1)^3]=3-2[8+(-1)] \\ \\ =3-2(8-1)=3-2(7)=3-14 \\ \\ -11](https://tex.z-dn.net/?f=3-2%5B2%5E3%2B%28-1%29%5E3%5D%3D3-2%5B8%2B%28-1%29%5D%20%5C%5C%20%20%5C%5C%20%3D3-2%288-1%29%3D3-2%287%29%3D3-14%20%5C%5C%20%20%5C%5C%20-11)
Part 12
Part 2
Part 3
Part 4
Part 5
Part 6
Part 7
Part 8
Part 9
Part 10
Part 11
Part 12
is 392 a perfect cube? if not what is the what is the smallest natural number by which 392 must be multiplicated so that the pro
1 answer:
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Answer:
George is 43.20 ft East of his starting point.
Step-by-step explanation:
Let Paula's speed be x ft/s
George's speed = 9 ft/s
Note that speed = (distance)/(time)
Distance = (speed) × (time)
George takes 50 s to run a lap of the track at a speed of y ft/s
Meaning that the length of the circular track = y × 50 = 50y ft
George and Paula meet 14 seconds after the start of the run.
Distance covered by George in 14 seconds = 9 × 14 = 126 ft
Distance covered by Paula in 14 seconds = y × 14 = 14y ft
But the sum of the distance covered by both runners in the 14 s before they first meet each other is equal to the length of the circular track
That is,
126 + 14y = 50y
50y - 14y = 126
36y = 126
y = (126/36) = 3.5 ft/s.
Hence, Paula's speed = 3.5 ft/s
Length of the circular track = 50y = 50 × 3.5 = 175 ft
So, in 4 minutes (240 s), with George running at 9 ft/s, he would have ran a total distance of
9 × 240 = 2160 ft.
2160 ft around a circular track of length 175 ft, means that George would have ran a total number of laps (2160/175) = 12.343 laps.
Breaking this into 12 laps and 0.343 of a lap from the starting point. 0.343 of a lap = 0.343 × 175 = 60 ft
So, 60 ft along a circular track subtends an angle θ at the centre of the circle.
Length of an arc = (θ/360°) × 2πr
2πr = total length of the circular track = 175
r = (175/2π) = 27.85 ft
Length of an arc = (θ/360) × 2πr
60 = (θ/360°) × 175
(θ/360°) = (60/175) = 0.343
θ = 0.343 × 360° = 123.45°
The image of this incomplete lap is shown in the attached image,
The distance of George from his starting point along the centre of the circular track = (r + a)
But, a can be obtained using trigonometric relations.
Cos 56.55° = (a/r) = (a/27.85)
a = 27.85 cos 56.55° = 15.35 ft
r + a = 27.85 + 15.35 = 43.20 ft.
Hence, George is 43.20 ft East of his starting point.
Hope this Helps!!!
