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svp [43]
3 years ago
5

What is the length of AB when A is (-7,-6) and B is (1,0)

Mathematics
1 answer:
NemiM [27]3 years ago
5 0

Answer:

The answer is

<h2>10 units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

d =  \sqrt{( {x1 - x2})^{2}  +  ({y1 - y2})^{2} }  \\

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

A (-7,-6) and B (1,0)

The distance between them is

|AB|  =  \sqrt{( { - 7 - 1})^{2}  +  ({ - 6 - 0})^{2} }  \\  =  \sqrt{ ({ - 8})^{2}  +  ({ - 6})^{2} }   \\  =  \sqrt{64 + 36}  \\  =  \sqrt{100}  \:  \:  \:  \:  \:  \:  \:  \:

We have the final answer as

<h3>10 units</h3>

Hope this helps you

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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
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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

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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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The random variable X has the following probability density function: fX(x) = ( xe−x , if x &gt; 0 0, otherwise. (a) Find the mo
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Step-by-step explanation:

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