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Yuri [45]
3 years ago
13

Simplify (3+√5)(5+√3)

Mathematics
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer:

answer of this question is

35.24

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I can create and use a graph to predict values and justify my results? Help me with this please I would appreciate it
mestny [16]

Answer:

simple even tho im in 6th grade it would be over $100

5 0
3 years ago
George and Paula are running around a circular track. George starts at the westernmost point of the track, and Paula starts at t
arsen [322]

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!!!

6 0
3 years ago
On a windy day, a 90 foot rope tightly secures a hot air balloon to a stake in the ground. From the balloon, the angle of depres
Alborosie

Answer:

  • 79 ft

Step-by-step explanation:

Balloon - B, stake - S, point at ground - G

<u>Given:</u>

  • BS = 90 ft
  • m∠BSG = 62°
  • BG = x

<u>Use sine and solve for x:</u>

  • BG/BS = sin ∠BSG
  • sin 62° = x / 90
  • x = 90 sin 62 deg
  • x = 79 ft (rounded)
4 0
3 years ago
Two similar trapezoids have areas
puteri [66]
It d i hope this helps
3 0
2 years ago
Read 2 more answers
Seventh grade
aleksandrvk [35]

Answer:

2.8 hours

Step-by-step explanation:

36.68 / 13.1 = 2.8

7 0
2 years ago
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