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zhannawk [14.2K]
2 years ago
11

The populations of two cities after t years can be modeled by -150t+50,000 and 50t+75,000 . What is the difference in the popula

tions of the cities when t=4 ?
Mathematics
2 answers:
Alex2 years ago
4 0
50t+34 t=4 is accelerating
creativ13 [48]2 years ago
3 0

Answer:

The difference in the populations of the cities when t=4 is 25,800.

Step-by-step explanation:

Given equations each of the city:

-150t+50,000

and

50t+75,000

To find the difference between the population when t=4, we first have to solve both equation when t=4.

For both equations, substitute 4 into t:

-150(4) + 50,000 = 49,400

50(4) + 75,000 = 75,200

Now subtract 49,400 from 75,200 and you get 25,800.

The difference in the populations of the cities when t=4 is 25,800.

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3 years ago
Ruth has a beaker containing a solution of 800 mL of acid and 200 mL of water. She thinks the solution is a little strong, so sh
NISA [10]

Answer:

350 mL of water

Step-by-step explanation:

Well she starts with 200mL of water and there is 800 mL of acid of water.

She drains 100 mL of acid and adds 100 mL of water so there is 300 mL of water.

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Then she drains 100 mL and she they are mixed she drains half of acid and half of water so she has 250 mL of water.

The she adds 100 mL of water so now there’s 350 mL of water left.

3 0
3 years ago
Round 3292.53429 to the nearest hundred thousandth
kari74 [83]

Answer:

The answer is "0.00001"

Step-by-step explanation:

Given value:

\to 3292.53429

when we round a hundred thousandths (100,000) value. So, the given rounding 3292.53429 to the nearest 0.00001.

7 0
3 years ago
Find the distance between the lines<br> Y=3x+10<br> Y=3x-20<br> AND <br> Y=3/2x+3/2<br> Y=3/2x-5
rosijanka [135]
At first we need to know how to find the distance between two parallel lines
The parallel lines which have the same slope 
If we have two parallel lines ⇒ y₁ = mx + c₁   and    y₂ = mx + c₂
So, the distance between y₁ and y₂ = d = \frac{|c_{1} - c_{2}|}{\sqrt{1 + m^{2}}}
========================================================

Part (1):
=======

<span>The distance between the lines
</span>
<span> Y₁ = 3x + 10    and       Y₂ = 3x - 20</span>
∴ m = 3  , c₁ = 10  and  c₂ = -20
∴ d = \frac{| 10 - (-20)|}{\sqrt{1 + 3^{2}}}= \framebox{9.5}

========================================================

Part (2):
=======

<span>The distance between the lines
</span>
Y₁ = <span>3/2 x + 3/2</span>    and       Y₂ = <span>3/2 x - 5</span>
∴ m = 3/2 = 1.5  , c₁ = 3/2 = 1.5   and  c₂ = -5
∴ d = \frac{| 1.5- (-5)|}{\sqrt{1 + 1.5^{2}}}= \framebox{3.6}

6 0
3 years ago
Two boats sail away from a buoy making a 90 degree angle. The first boat is 165 meters away and is traveling at 25 m/s. The seco
NikAS [45]

Answer:


Step-by-step explanation:

lets say in the shown triangle x=165

so speed \frac{dx}{dt} =25

and  \frac{dy}{dt} =35


Also since distance = speed* time

time = pi minutes = pi/60 seconds

y = \frac{35\pi \pi }{60}

y = \frac{7\pi }{12}


Now using the pythagorean :

x^2+y^2=z^2\\so \\165^2+(\frac{7\pi }{12} )^2 =z^2

<u>derivate the equation so  we get :</u>

x\frac{dx}{dt} +y\frac{dy}{dt} =z\frac{dz}{dt} \\165*25+35*\frac{7\pi }{12}=\sqrt{165^2+(\frac{7\pi }{12})^2} *\frac{dz}{dt}


\frac{dz}{dt} =25.387 m/s

So Rate of change = <u><em>25.387 m/s</em></u>


5 0
3 years ago
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