Geometry math question

Mathematics

1 answer:

Brut [27]4 years ago

7 0

Look at the picture.

Let |AE| = |AD| = b

We have a proportion:

$\dfrac{x}{13}{y}=\dfrac{19}{y+a}=\dfrac{x}{y+2a}$

Solve for y from first proportion

$\dfrac{13}{y}=\dfrac{19}{y+a}\ \ \ |\text{cross multiply}\\\\19y=13(y+a)\\\\19y=13y+13a\ \ \ \ |-13y\\\\6y=13a\ \ \ |:6\\\\y=\dfrac{13}{6}a$

Substitute to the second proportion

$\dfrac{19}{y+a}=\dfrac{x}{y+2a}\\\downarrow\\\dfrac{19}{\frac{13}{6}a+a}=\dfrac{x}{\frac{13}{6}a+2a}\\\\\dfrac{19}{\frac{13}{6}a+\frac{6}{6}a}=\dfrac{x}{\frac{13}{6}a+\frac{12}{6}a}\\\\\dfrac{19}{\frac{19}{6}a}=\dfrac{x}{\frac{25}{6}a}\ \ \ \ |\dot a\neq0\\\\\dfrac{19}{\frac{19}{6}}=\dfrac{x}{\frac{25}{6}}\\\\19\cdot\dfrac{6}{19}=x\cdot\dfrac{6}{25}\\\\6=\dfrac{6x}{25}\ \ \ |\cdot25\\\\6x=6\cdot25\ \ \ |:6\\\\x=25$

Answer: B. 25

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A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

Novosadov [1.4K]

Answer:

$A(t)=240-220e^{-\frac{t}{40}}$

Step-by-step explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

Volume of the tank = 240 liters
Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

$R_{in}=$ (concentration of salt in inflow)(input rate of fluid)

$R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}$

$R_{out}=$ (concentration of salt in outflow)(output rate of fluid)

$R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}$

Rate of change of the amount of salt in the tank:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{40}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{40}=6\\$The integrating factor: e^{\int \frac{1}{40}dt} =e^{\frac{t}{40}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{40}}+\dfrac{A}{40}e^{\frac{t}{40}}=6e^{\frac{t}{40}}\\(Ae^{\frac{t}{40}})'=6e^{\frac{t}{40}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{40}})'=\int 6e^{\frac{t}{40}} dt\\Ae^{\frac{t}{40}}=6*40e^{\frac{t}{40}}+C, $(C a constant of integration)\\Ae^{\frac{t}{40}}=240e^{\frac{t}{40}}+C\\$Divide all through by e^{\frac{t}{40}}\\A(t)=240+Ce^{-\frac{t}{40}}$