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3 years ago

¿Cuál es la ecuación de la recta que pasa por el punto? (-1,-5) y tiene una pendiente de - 1?

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

6 0

Answer:

y=-1x-6

Step-by-step explanation:

y-y1=m(x-x1)

y-(-5)=-1(x-(-1)

y+5=-1(x+1)

y+5=-1x-1

y=-1x-5-1

y=-1x-6

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A 250-gal tank is initially filled with brine (salt-water mixture) containing 100 lb of salt. Brine containing 3 lb of salt per

gulaghasi [49]

Answer:

x(t) = 581.66 lb

Step-by-step explanation:

From the given information:

Consider the salt quantity in the tank at time t be x(t) lb.

∴ the rate of change of salt content in the tank is $\dfrac{dx}{dt}= Rate \ of \ \ inflow - rate \ of \ \ outflow$

where:

the rate of inflow = salt conc. × flow rate = 3 lb/gallon × 7 gallons

=21 lb/s

rate of outflow = salt conc. in the tank × flow rate

= x/250 × 9

= 9x/250 lb/s

∴

$\dfrac{dx}{dt} = 21 - \dfrac{9x}{250}$

$\dfrac{dx}{dt} + \dfrac{9x}{250}= 21$

This is a 1st order linear differentiation,

The integrating factor if $e^{^{ \int \dfrac{9}{250}\ dt}} = e^{^{ \dfrac{9t}{250} }}$

∴

$e^{^{ \dfrac{9t }{250}}} \ \ x(t) = \int 21 e^{\dfrac{9t}{250}} \ dt + C$

$e^{^{ \dfrac{9t }{250}}} \ \ x(t) = 21 \times \dfrac{250}{9}e^{\dfrac{9t}{250}} + C$

$x(t) = \dfrac{1750}{3}+Ce^{^{\dfrac{-9t}{250}}}$ at t = (0) and x(0) = 100 lb

Hence;

$100 = \dfrac{1750}{3}+C$

$C = \dfrac{1750}{3} -100$

$C = - \dfrac{1450}{3}$

∴ $x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-9t}{250}}}$

after time t = 1 minute i.e 60 s

$x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-9 \times 60}{250}}}$

$x(t) = \dfrac{1750}{3}-\dfrac{1450}{100}e^{^{\dfrac{-540}{250}}}$

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3 years ago

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vichka [17]

Answer:

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Step-by-step explanation:

the equation of a line is y = ax+ b where a is the slope, b is the y-intercept

let us take yB - yA

→ yB - yA = axB + b - axA - b

→ yB - yA = a(xB - xA)

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2 years ago

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yawa3891 [41]

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3 years ago

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mixas84 [53]

Answer:

C.) Solve for the varible

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2 years ago

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