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saveliy_v [14]
1 year ago
12

Hi, I actually have this question not this one sorry

Mathematics
1 answer:
Ivanshal [37]1 year ago
4 0

we know that

If V is the midpoint of SU

then

SV=VU

substitute the given values

2x+18=8x-6

solve for x

8x-2x=18+6

6x=24

<h2>x=4</h2>

Find SV

SV=2x+18

SV=2(4)+18

<h2>SV=26 units</h2>

so

<h2>VU=26 units</h2>

and

SU=2SV

SU=2(26)

<h2>SU=52 units</h2>
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shutvik [7]

Answer:

a: y=70x

b: y=70x+100

c: They are parallel to each other, with equation b's line being 100 dollars above equation a.

Step-by-step explanation:

They both have the same slope (70), but new customers will have a y intercept of 100 because of the one-time fee.

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3 years ago
Read 2 more answers
Initially tank I contains 100 litres of salt brine with a concentration of 1 kilogram per litre, and tank II contains 100 litres
Gala2k [10]

Answer:

a)A1(t)=\frac{100000000}{(100-t)(100+t)^{2} } \\C1(t)=\frac{A1(t)}{100+t}

b) C1 = 0.8348 [kg/lt]

Step-by-step explanation:

Explanation

First of all, the rate of change of the amount of salt in the tank I is equal to the rate of change of salt incoming less the rate change of the salt leaving, so:

\frac{dA1(t)}{dt}= R_{in}C_{in}-R_{out}C_{out}

We know that the incoming rate is greater than the leaving rate, this means that the fluid in the tank I enters more than It comes out, so the total rate is :

R_{total}=R_{in}-R_{out}=\frac{2 lt}{min} - \frac{1 lt}{min}=  \frac{1 lt}{min}

This total rate means that 1 lt of fluid enters each minute to the tank I from the tank II, with the total rate we can calculate the volume in the tank I y tank II as:

V_{I}=100 lt + Volumen_{in}=  100 lt + (\frac{1lt}{min})(t) =100+t

V_{II}=100 lt - Volumen_{out}=  100 lt - (\frac{1lt}{min})(t) =100-t

Now we have the volume of both tanks, the next step is to calculate the incoming and leaving concentration. The concentration is the ratio between the amount of salt and the volume, so:

C(t)=C_{out} =\frac{A1(t)}{V_{I} }=\frac{A1(t)}{100+t }

Since fluid is pumped from tank I into tank II, the concentration of the tank II is a function of the amount of salt of the tank I that enters into the tank II, thus:

C_{in} =\frac{(A1(t)/V_{I})(t)}{V_{II} }=\frac{A1(t)}{V_{I} V_{II}}(t)

C_{in} =\frac{A1(t)}{(100+t)(100-t)}(t)=\frac{A1(t)}{(10000-t^{2} )}(t)

If we substitute the concentrations and the rates into the differential equation we can get:

\frac{dA1(t)}{dt}= R_{in}C_{in}-R_{out}C_{out}\\\frac{dA1(t)}{dt}= (2)(\frac{(t)A1(t)}{10000-t^{2} })-(1)(\frac{(A1(t)}{100+t })

\frac{dA1(t)}{dt}= A1(t)(\frac{2t}{10000-t^{2} }-\frac{1}{100+t })

\frac{dA1(t)}{dt}- (\frac{2t}{10000-t^{2} }-\frac{1}{100+t })A1(t)=0

The obtained equation is a homogeneous differential equation of first order and the solution is:

a) A1(t)= \frac{100000000}{(100-t)(100+t)^{2} }

and the concentration is:

C1(t)= \frac{100000000}{(100-t)(100+t)^{3}}

This equations A1(t) and C1(t) are only valid to 0<=t<100 because to t >=100 minutes the tank II will be empty and mathematically A1(t>=100) tends to the infinite.

b) To calculate the concentration in the tank I after 10 minutes we have to substitute t=10 in C1(t), thus:

C1(10)= \frac{100000000}{(100-10)(100+10)^{3}}=0.8348 kg/lt

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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It appears that we cannot use the commutative property on this expression. But we can. First, we need to change - 5 = + -5.

7x - 5 = 7x + -5

Because we now have addition, we can prove the commutative property.

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Answer: $52.50

Step-by-step explanation:

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