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lidiya [134]
3 years ago
7

in the diagram below,RQ is parallel to ST. ST =4cm RP =10cm PT = 5cm a.find,. i.the scale factor. ii.|RQ|. b.if the area of ∆PRQ

is 80 cm find the area of triangle ∆PST.​

Mathematics
1 answer:
den301095 [7]3 years ago
5 0

Answer:

  1. us7e7eoeod8rr8r88rr8 fnddkdkiduduru
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Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
aliya0001 [1]

Answer:

A=1500-1450e^{-\dfrac{t}{250}}

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

R_{in} =(concentration of salt in inflow)(input rate of brine)

=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}

R_{out}=(concentration of salt in outflow)(output rate of brine)

=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}

Now, the rate of change of the amount of salt in the tank

\dfrac{dA}{dt}=R_{in}-R_{out}

\dfrac{dA}{dt}=6-\dfrac{A}{250}

We solve the resulting differential equation by separation of variables.  

\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}

Taking the integral of both sides

\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}

Recall that when t=0, A(t)=50 (our initial condition)

50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}

4 0
3 years ago
CD is the segment bisector of AB at D. If AD = 4x-1 and DB = 9x-21. what is the length of AD and AB
Artemon [7]

Answer:

AD= 15

AB=30

Step-by-step explanation:

We are given that the Point D bisect the line segment AB

Hence AD = DB

AD= 4x-1

BD=9x-21

Hence

4x-1=9x-21

adding 21 on both sides and subtracting 4x from both sides

4x-4x-1+21=9x-4x-21+21

-1+21=5x

5x=20

dividing both sides by x we get

x=4

Hence

AD = 4x-1

= 4(4)-1= 16-1=15

AD= 15

AB = 2 x AD

     = 2 x 15

  = 30

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4 years ago
Simplify.<br><br><br><br> Enter your answer, in simplest radical form, in the box.
puteri [66]
24.75 rounded might be 25
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What is the equation for the line of reflection?<br><br> x=6 <br> y=6<br> y=x<br> y=2
shepuryov [24]

Answer:

Hi there!

The answer to this question is: x=6

Step-by-step explanation:

All the points are being mirrored across the line x=6.

7 0
4 years ago
Find the ​twenty-ninthninth term of the arithmetic sequence 22​, 20​, 18​, ...
soldi70 [24.7K]
The Twenty-Ninth term is -34. The arithmetic sequence is degrading by -2. Below is the sequence.

22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2, 0, -2, -4, -6, -8, -10, -12, -14, -16, -18, -20, -22, -24, -26, -28, -30, -32, -34
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3 years ago
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