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Helen [10]
3 years ago
7

Is (2,3) (4,5) (2,8) a function?

Mathematics
2 answers:
snow_lady [41]3 years ago
4 0

Answer:

No

Step-by-step explanation:

if the X number ever repeats, then its never a function

lukranit [14]3 years ago
3 0
Answer:
This is not a function.

Explanation:
Since x=2 produces y=3 and y=8 the relation is not a function.

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Answer:    13.7777777778

Step-by-step explanation:

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Kaiya says the complex fraction /= is equivalent to the complex fraction =/. Is she correct? Explain your answer.
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Answer:

Well i think so

Step-by-step explanation:

its just an equal sign on different sides

6 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}}

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3 years ago
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B 6 whole peices

explanation
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3 years ago
write the slope intercept equation for the line that passes through (-8,11) and is perpendicular t 2x -7y=14
SIZIF [17.4K]
General\ formula:\\y=ax+b\\\\2x-7y=14\\-7y=14-2x\\y=\frac{2}{7}x-2\\\\Line\ is\ perpendicular&#10; when\ a=-\frac{7}{2}\\y=-\frac{7}{2}x+b\\\\Passes\ through\ (-8,11):\\11=(-\frac{7}{2})*(-8)}+b\\11-28=b\\b=-17\\\\y=\frac{7}{2}-17
4 0
4 years ago
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