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3 years ago

7

What is expanded form 12.38

2 answers:

Aleks [24]3 years ago

6 0

Expanded form = Twelve and thirty- eight hundredths.<span>
</span>

Kruka [31]3 years ago

5 0

Twelve and thirty- eight hundredths.

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What is the slope of this line?<br><br> (13,9)

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Answer:

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Step-by-step explanation:

Whenever the equation of a line is written in the form y = mx + b, it is called the slope-intercept form of the equation. The m is the slope of the line. And b is the b in the point that is the y-intercept (0, b). For example, for the equation y = 3x – 7, the slope is 3, and the y-intercept is (0, −7).

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3 years ago

What is the slope of the line that passes through the points (9, 1) and (17, 2)

Wewaii [24]

Answer:

C

Step-by-step explanation:

Slope of a line that goes trough the points (x1=9, y1=1) and (x2=17, y2=2)

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2 years ago

A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

$\frac{dC}{dt}=Ci*Qi-Co*Qo$

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

$\frac{dC}{dt}=-Co*Qo$

Rearranging the equation, it becomes

$\frac{dC}{C}=-Qo*dt$

Integrating both sides

$\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}$

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

$C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\$

The final equation for the concentration of salt at any given time is

$C=exp^{-3*t-0.693}$

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

$C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}= -\frac{-0.266}{3}=0.088$

5 0

3 years ago