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PIT_PIT [208]
3 years ago
7

What is 62x37 I really need some help

Mathematics
2 answers:
iragen [17]3 years ago
7 0

Answer:

2294

Step-by-step explanation:

Marysya12 [62]3 years ago
5 0

Answer:

2294

Step-by-step explanation:

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Which system of linear inequalities is represented by the
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first choice: y≥1/3 x+3 and 3x-y>2

Step-by-step explanation:

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Try This question and I’ll give you brainliest
Amanda [17]

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3

Step-by-step explanation:

We want to cancel out the 3y.  The second equation has a -y so if we multiply that by 3 we'll get -3y and when we add the ys go away.

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Please simplify this expression thanks
Kruka [31]

Answer:

Step-by-step explanation:

Factor numerator: x^2 + x - 12 = (x + 4)(x - 3)

Factor denominator: x^2 + 7x + 12 = (x + 3)(x + 4)

Notice that the numerator and denominator have a common factor (x + 4).

They can cancel.

What is left is (x - 3)/(x + 3)

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3 years ago
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kondor19780726 [428]
The answer is 16, just put the numbers in
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3 years ago
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the air in a small room 12ft by 8ft by 8ft is 3% carbon monoxide. Starting at t=0, fresh air containing no carbon monoxide is bl
Irina-Kira [14]

Answer:

The air in the room at 0.01% carbon monoxide at 43.8 min

Step-by-step explanation:

Let be the volume of CO in the room at time t, be v(t)  and  the total volume of the room be V. The volume percent of CO in the room at a given time is then given by:

p(t) = \frac{100\times v(t)}{V}

Volume percent is the measure of concentration used in this problem. The "Amount" of CO in the room is then measured in terms of the volume of CO in the room.

Let the rate at which fresh air enters the room be f, which is the same as the rate at which air exits the room. We assume that the air in the room mixes instantaneously with the air entering the room, so that the concentration of CO is uniform throughout the room.

As you wrote, the rate at which the volume of CO in the room changes with time is given by

\frac{dvt}{dt} = 0 \times f -\frac{f}{v} \times v(t) = -\frac{f}{v} \times v(t)

This is a simple first-order equation:

\frac{dv}{v} = -\frac{f}{v} dt

ln(v) - ln(c) = -\frac{f}{v} \times t

where ln(c) is the constant of integration.

ln \frac{v}{c} = -\frac{f}{v} \times t

v(t) = c \times e^{(-f*\frac{t}{V})}

In terms of volume percent,

p(t) = \frac{100*v(t)}{V}= (\frac{C}{V})*exp(\frac{-f \times t}{v})

where C = 100*\frac{c}{V} is just another way of writing the constant.

Plugging in the values for the constants, we get:

p(t) = (\frac{C}{768 cu.ft.})* exp(\frac{-t}{7.68 min})

Now use the initial condition (p(0) = 3% at t = 0) to solve for C:

3% = C

p(t) = (3\%)\times exp(\frac{t}{7.68 min})

To find the time when the air in the room reaches a certain value, it is easier to rewrite this solution as:

\frac{p(t)}{3\%} = exp(\frac{-t}{7.68 min})

t(p) = -(7.68 min)ln(\frac{p}{3\%} )

= (7.68 min)*ln(\frac{3\%}{p})

The question asks when p(t) = 0.01%. Plugging this into the above equation, we get:

t(0.01\%) = (7.68 min)*ln(\frac{3}{0.01}) = 43.8 min

3 0
3 years ago
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