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Andre45 [30]
2 years ago
10

Help me! pls, i need to get rid of points :) say hey and ill give you some!

Mathematics
2 answers:
timurjin [86]2 years ago
8 0

Answer:

did she reallly make you ?

Step-by-step explanation:

one your ment have been birthed

earnstyle [38]2 years ago
5 0
Heyyyyyyyyyyyyyyyyyyyyyyyyy
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An Amtrak train travels at 125 mile per hour. Convert the speed to miles per minute round to the nearest tenth
Tomtit [17]
To convert this you divide the speed by 60 since there's 60 minutes in an hour
125/60 = 2.0833....

Since you round the answer would be
2.1 miles per minute
6 0
3 years ago
Read 2 more answers
The circle below represents one whole <br><br> What percent is represented the shaded area
allsm [11]

Answer:

50%

Step-by-step explanation:

Imagine a pizza, the pizza is cut into 4 slices and 2 have already been eaten. There you go! You now have the fraction 2/4. 2/4=1/2. 1/2 as a percentage is 50%.

8 0
3 years ago
What is the slope of the line represented by the equation y = –2/3 – 5x
Nookie1986 [14]
The equation is of the form y = mx + c, where m is the slope and c just a constant. In this example, m=-5 and c=-2/3, so the slope is -5. The -2/3 is just a vertical offset and is irrelevant.
6 0
3 years ago
Can you help me solve this problem<br> |2x-2|&gt;8
allsm [11]

2x-2>8

2x>10

x = 10/2 = 5

x>5

 since it is an absolute equation

it could also be x<-3

 answer is x>5 or x<-3

7 0
3 years ago
A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/mi
sergij07 [2.7K]

Answer:

(a) After time (t), the amount of salt in the tank is s(t) = 130 − 130e ∧−3t/200 / 3

(b) After 60 minutes, the salt in the tank is s(60) ≈ 25.7153

Step-by-step explanation:

To start with,

Let s(t) = amount of salt in kg of salt at time t.

Then we have:

dt/ds =  (rate of salt into tank) − (rate of salt out of tank)

= (0.05 kg/L · 5 L/min) + (0.04 kg/L · 10 L/min) − ( s kg/ 1000 L x 15 L/min)

= 0.25 kg/min + 0.4 kg/min − 15s / 1000 kg/min

So we get the differential equation

dt/ds = 0.65 − 15s / 1000

= 65 / 100 − 15s / 1000

dt/ds = 130 - 3s / 200

We separate s and t to get

1 / 130 - 3s ds = 1 / 200 dt

Then we Integrate,

Thus we have

∫1 / 130 - 3s ds = ∫1 / 200 dt

- 1/3  · ln |130 − 3s| =1 / 200 t + C1

ln |130 − 3s| = - 3 /200 t + C2

|130 − 3s| = e− ³⁺²⁰⁰ ∧ t+C2

|130 − 3s| = C3e − ∧3t/200

130 − 3s = C4e  ∧−3t/200

−3s = −130 + C4e  ∧−3t/200

s = 130 - C4e ∧−3t/200 / 3

Since we begin with pure water, we have s(0) = 0. Substituting,

0 = 130 − C4e  ∧−3·0/200 / 3

0 = 130 − C4

C4 = 130

So our function is

s(t) = 130 − 130e ∧−3t/200 / 3

After one hour (60 min), we have

s(60) = 130 − 130e ∧−3·60/200 / 3

s(60) ≈ 25.7153

Thus, after one hour there is about 25.72 kg of salt in the tank.

6 0
3 years ago
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