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2 years ago

Please help me with this, corresponding parts

Mathematics

1 answer:

poizon [28]2 years ago

4 0

Answer:

∠ C = 58°

Step-by-step explanation:

DB is a perpendicular bisector , so

∠ ADC = 2 × 32° = 64°

Since AD = CD then Δ ACD is isosceles with the base angle congruent , then

∠ C = $\frac{180-64}{2}$ = $\frac{116}{2}$ = 58°

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3 years ago

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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.

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

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:

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

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3 years ago

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uysha [10]

I'm not sure how you obtained -5 as your answer but this is the mathematical explanation. hope this helps!
${x}^{2} + 3x = 40 \\ {x}^{2} + 3x - 40 = 0 \\(x + 8)(x - 5) = 0 \\ x = - 8 \\ x = 5$

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3 years ago