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o-na [289]
3 years ago
6

Building a is 271 meters taller than building

Mathematics
1 answer:
kvv77 [185]3 years ago
4 0

Let a and b represent the heights of the corresponding buildings (in meters).

... a = b +271 . . . . . . . a is 271 meters taller than b

... 2b -a = 211 . . . . . . if a is subtracted from twice b, the result is 211

Use the expression for a in the first equation to substitute for a in the second.

... 2b - (b+271) = 211

... b = 482 . . . . . . . . . . . simplify and add 271

... a = b +271 = 753

Building a is 753 meters tall; building b is 482 meters tall.

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Step-by-step explanation:

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Evgen [1.6K]

When x = 0 the value of the function is 2 so its either A or B.

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If 3 = 1/2gt^2, make t the subject of the formula<br>​
Montano1993 [528]

Answer:

t = √1.5/g

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5 0
3 years ago
When running a 100 meter race Bill reaches his maximum speed when he is 45 meters from the starting line and 5 seconds have elap
Mkey [24]

Bill's speed is constant after 5 seconds and 45 meters into the race.

Correct responses:

a. Bill's maximum speed is 8 m/s

b. 24 meters

c. 57 meters

<h3>Methods used to calculate speed and distance traveled</h3>

Given parameters are;

The distance of the race = 100 meter

Distance at which Bill reaches maximum speed = 45 meters

Speed Bill maintains after 5 seconds = The maximum speed

Time at which Bill is 85 meters from the starting line = 10 seconds after start

a. Required;

Bill's maximum speed in meters.

Solution:

Distance Bill runs at maximum speed, d = 85 m - 45 m = 40 m

Time at which Bill runs the 40 m at maximum speed, <em>t</em> = 10 s - 5 s = 5 s

Speed = \mathbf{\dfrac{Distance}{Time}}

Therefore;

  • Bill's \  maximum \ speed = \dfrac{40 \ m}{5 \ s} = \underline{8 \ m/s}

b. i. Required:

The distance Bill will run for 3 seconds at the maximum speed;

Solution:

Distance,<em> s</em> = Speed, <em>v</em> × Time, <em>t</em>

<em />

The distance traveled at maximum speed in 3 seconds is therefore;

Distance = 8 m/s × 3 s = 24 m

The distance Bill travels in 3 seconds at the maximum speed is 24 meters.

ii) The distance Bill travels at 8 seconds after start, is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 3 seconds = 24 meters

Therefore;

  • Bill's distance from the starting line 8 seconds after start is 45 meters + 25 meters = <u>69 meters</u>

c. Bill's distance 6.5 seconds after the start of the race is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 1.5 seconds = 1.5 s × 8 m/s = 12 meters

  • Bill's distance from the starting line 6.5 seconds after start of the race = 45 meters + 12 meters = <u>57 meters</u>

Learn more about distance, speed, time, relationship here:

https://brainly.in/question/49075584

6 0
2 years ago
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