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Mashutka [201]
3 years ago
5

A ratio is a comparison of two numbers by addition. always sometimes never

Mathematics
2 answers:
solong [7]3 years ago
6 0

The answer would be the Third Option, "Never".

horsena [70]3 years ago
6 0

A ratio is never a comparison of two numbers by addition. [never]

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Pleaseee help me w dis asap!!
ikadub [295]

Answer:

x=2 f(x)=5-x

o≤x≤3  f(x)=x

2<x<3 f(x)=1

3<x≤5  f(x)=5-x

Step-by-step explanation:

6 0
3 years ago
Draw an ordered stem and leaf diagram to show this information.
IceJOKER [234]

Answer:

24|5

25|0;1;3;8

26|4;7;9

27|3;6

Step-by-step explanation:

key in the bottom right:

24|5means 245

5 0
3 years ago
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If 0 is a positive acute angle and 2 cos 0+3 = 4,<br> find the number of degrees in 0.
Lemur [1.5K]
Answer: 1 0 is like x you solve for x. X= 1 0=1
4 0
3 years ago
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
What is a common denominator for 1/6 and 3/5
d1i1m1o1n [39]

Answer:

30

Step-by-step explanation:

8 0
3 years ago
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