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3 years ago

Yesterday the temperature at noon was 12.2°F. By midnight, it had dropped by 20.9°F. What was the temperature at midnight?

Mathematics

1 answer:

Gekata [30.6K]3 years ago

5 0

Answer:

The temperature dropped down to -8.7°F

Step-by-step explanation:

12.2°F - 20.9°F = -8.7°F

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Jacky baked some biscuits.he kept 3/7 of the biscuits in container A.5/8 of the remainder in container B and rest in container C

Colt1911 [192]

Let "b" represent the number of biscuits Jacky baked.
.. container A holds (3/7)b
.. container B holds (1 -3/7)*(5/8)b = (5/14)b
.. container C holds (1 -3/7 -5/14)b = (3/14)b = 168

Then
.. b = 168/(3/14) = 784

Jacky baked 784 biscuits.

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3 years ago

(1 point) A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of

Paraphin [41]

1. dy/dt is the net rate of change of salt in the tank over time. As such, it's equal to the difference in the rates at which salt enters and leaves the tank.

The inflow rate is

(0.4 kg/L) (6 L/min) = 2.4 kg/min

and the outflow rate is

(concentration of salt at time t) (4 L/min)

The concentration of salt is the amount of salt (in kg) per unit volume (in L). At any time t > 0, the volume of solution in the tank is

100 L + (6 L/min - 4 L/min) t = 100 L + (2 L/min) t

That is, the tank starts with 100 L of pure water, and every minute 6 L of solution flows in and 4 L is drained, so there's a net inflow of 2 L of solution per minute. The amount of salt at time t is simply y(t). So, the outflow rate is

(y(t)/(100 + 2t) kg/L) (4 L/min) = 2 y(t) / (50 + t) kg/min

and the differential equation for this situation is

$\dfrac{dy}{dt} = 2.4 \dfrac{\rm kg}{\rm min} - \dfrac{2y}{50+t} \dfrac{\rm kg}{\rm min}$

There's no salt in the tank at the start, so y(0) = 0.

2. Solve the ODE. It's linear, so you can use the integrating factor method.

$\dfrac{dy}{dt} = 2.4 - \dfrac{2y}{50+t}$

$\dfrac{dy}{dt} + \dfrac{2}{50+t} y = 2.4$

The integrating factor is

$\mu = \displaystyle \exp\left(\int \frac{2}{50+t} \, dt\right) = \exp\left(2\ln|50+t|\right) = (50+t)^2$

Multiply both sides of the ODE by µ :

$(50+t)^2 \dfrac{dy}{dt} + 2(50+t) y = 2.4 (50+t)^2$

The left side is the derivative of a product:

$\dfrac{d}{dt}\left[(50+t)^2 y\right] = 2.4 (50+t)^2$

Integrate both sides with respect to t :

$\displaystyle \int \dfrac{d}{dt}\left[(50+t)^2 y\right] \, dt = \int 2.4 (50+t)^2 \, dt$

$\displaystyle (50+t)^2 y = \frac{2.4}3 (50+t)^3 + C$

$\displaystyle y = 0.8 (50+t) + \frac{C}{(50+t)^2}$

Use the initial condition to solve for C :

$y(0) = 0 \implies 0 = 0.8 (50+0) + \dfrac{C}{(50+0)^2} \implies C = -100,000$

Then the amount of salt in the tank at time t is given by the function

$y(t) = 0.8 (50+t) - \dfrac{10^5}{(50+t)^2}$

so that after t = 50 min, the tank contains

$y(50) = 0.8 (50+50) - \dfrac{10^5}{(50+50)^2} = \boxed{70}$

kg of salt.

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2 years ago

Marika Perez’s gross biweekly pay is $2,768. Her earnings to date for the year total $80,272. What amount is deducted from her p

Angelina_Jolie [31]

Social Security $4,976.86 and for Medicare $1,163.94

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3 years ago

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The two-way table shows the number of students in a school who have bikes and/or skateboards at home:

Tpy6a [65]

If you are looking at the total of just people who have just bikes that's 40 and people who have just skateboards that's 10 so 30 but if its the total of bikes that's 72 and the total of skateboards that's 42 which is also 30. My best guess seeing as i don't know what math class you are in is 30.

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4 years ago

Find the original price. Discount is 30%. Sale price is $206.50

lakkis [162]

Your answer would be $295 :)

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3 years ago