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iogann1982 [59]
2 years ago
15

Which situation shows a constant rate of change?

Mathematics
1 answer:
kolbaska11 [484]2 years ago
3 0
The answer is most likely B. :)
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andriy [413]
60 plus 20 * 10= 200 so 200+60= 260 milligrams
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3 years ago
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avanturin [10]
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7 0
3 years ago
A farmer is able to prepare 1/2 are of soil in 3 days. How long will it take him to prepare 2 1/4 acres of soil?
Alexxx [7]

Answer:

x = 13 1/2 days

Step-by-step explanation:

1/2 acre              2 1/4 acre

------------ = -------------------------

3 days           x days


using cross products

1/2 * x = 2  1/4 * 3

change 2  1/4 to an improper fraction   (4*2+1)/4 =9/4

1/2 x = 9/4*3

1/2x = 27/4

multiply each side by 2

2 * 1/2 x = 27/4 * 2

x = 27/2

change back to a mixed number

x = 13 1/2

8 0
3 years ago
5c+16.5=13.5+10c<br> solve for c
lana66690 [7]

Answer:

c = 3/5

Step-by-step explanation:

5c+16.5=13.5+10c

Subtract 5c from both sides:

13.5+5c=16.5

5c=3

c = 3/5

7 0
3 years ago
Read 2 more answers
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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