All categories

3 years ago

Do you think mold is somthing we need to worry about in school

1 answer:

5 0

Personally, I think it is. Not only that mold is very unattractive and would not represent your school well, it can cause issues with your respiratory system. It can cause a lot of problems with children that have asthma and allergies or maybe they are sensitive to it. Even for the children who do not have those two, it can cause Illness and infections.

You might be interested in

A triangular road sign has a height of 3 feet and a base of 2.5 feet. How much larger in area is this sign than one with a heigh

jeka57 [31]

Answer:

Not larger or smaller; the areas are the same

Step-by-step explanation:

Recall that the formula for the area of a triangle of base b and height h is

A = (1/2)(b)(h).

In the first case, h = 3 ft and b = 2.5 ft. This results in an area of 3.75 ft^2.

In the second case, h = 2.5 ft and b = 3 ft. The area is the same: 3.75 ft^2.

This is because A = (1/2)(b)(h) has the same value regardless of the order in which b and h are used as multipliers.

There is no difference in the areas of the two signs. The areas are the same.

5 0

3 years ago

A rectangular prism has a length of 12 centimeters, a width of 6 centimeters, and a height of 9 centimeters. A cross section is

Klio2033 [76]

Answer:

<h2>The perimeter of the cross section is 30 centimeters.</h2>

Step-by-step explanation:

In this problem we have the intersection of a rectangular prism an a plane.

The dimensions of the rectangular plane are

$l=12cm\\w=6cm\\h=9cm$

Assuming the plane is cutting the prism horizontally, the cross section would have dimensions

$w=6cm\\l=9cm$

Because only the height would be cut.

So, the perimeter of the rectangle cross-section is

$P_{cross}=2(l+w)=2(6+9)=2(15)=30cm$

Therefore, the perimeter of the cross section is 30 centimeters.

3 0

3 years ago

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in the solution. Water containing1 lb

devlian [24]

Answer:

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is $\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$ .

(b) The concentration (in lbs per gallon) when it is at the point of overflowing is $\frac{121}{125}\:\frac{lb}{gal}$ .

(c) The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$ .

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x (concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x (concentration of substance in liquid exiting)

Let $C$ be the concentration of salt water solution in the tank (in $\frac{lb}{gal}$ ) and $t$ the time (in minutes).

Since the solution being pumped in has concentration $1 \:\frac{lb}{gal}$ and it is being pumped in at a rate of $3 \:\frac{gal}{min}$ , this tells us that the rate of the salt entering the tank is

$1 \:\frac{lb}{gal} \cdot 3 \:\frac{gal}{min}=3\:\frac{lb}{min}$

But this describes the amount of salt entering the system. We need the concentration. To get this, we need to divide the amount of salt entering the tank by the volume of water already in the tank.

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at t = 0 there were 200 gallons, 3 gallons are added and 2 are drained, and the net increase is 1 gallons per second. So,

$V(t)=200+t$

Therefore,

The rate at which $C(t)$ enters the tank is

$\frac{3}{200+t}$

The rate of the amount of salt leaving the tank is

$C\:\frac{lb}{gal} \cdot 2 \:\frac{gal}{min}+C\:\frac{lb}{gal} \cdot 1\:\frac{gal}{min}=3C\:\frac{lb}{min}$

and the rate at which $C(t)$ exits the tank is

$\frac{3C}{200+t}$

Plugging this information in the main equation, our differential equation model is:

$\frac{dC}{dt} =\frac{3}{200+t}-\frac{3C}{200+t}$

Since we are told that the tank starts out with 200 gal of solution, containing 100 lb of salt, the initial concentration is

$\frac{100 \:lb}{200 \:gal} =0.5\frac{\:lb}{\:gal}$

Next, we solve the initial value problem

$\frac{dC}{dt} =\frac{3-3C}{200+t}, \quad C(0)=\frac{1}{2}$

$\frac{dC}{dt} =\frac{3-3C}{200+t}\\\\\frac{dC}{3-3C} =\frac{dt}{200+t} \\\\\int \frac{dC}{3-3C} =\int\frac{dt}{200+t} \\\\-\frac{1}{3}\ln \left|3-3C\right|=\ln \left|200+t\right|+D\\\\$

We solve for C(t)

$C(t)=1+D(200+t)^{-3}$

D is the constant of integration, to find it we use the initial condition $C(0)=\frac{1}{2}$

$C(0)=1+D(200+0)^{-3}\\\frac{1}{2} =1+D(200+0)^{-3}\\D=-4000000$

So the concentration of the solution in the tank at any time t (before the tank overflows) is

$C(t)=1-4000000(200+t)^{-3}$

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is just the concentration of the solution times its volume

$(1-4000000(200+t)^{-3})(200+t)\\\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$

(b) Since the tank can hold 500 gallons, it will begin to overflow when the volume is exactly 500 gal. We noticed before that the volume of the solution at time t is $V(t)=200+t$ . Solving the equation

$200+t=500\\t=300$

tells us that the tank will begin to overflow at 300 minutes. Thus the concentration at that time is

$C(300)=1-4000000(200+300)^{-3}\\\\C(300)= \frac{121}{125}\:\frac{lb}{gal}$

(c) If the tank had infinite capacity the concentration would then converge to,

$\lim_{t \to \infty} C(t)= \lim_{t \to \infty} 1-4000000\left(200+t\right)^{-3}\\\\\lim _{t\to \infty \:}\left(1\right)-\lim _{t\to \infty \:}\left(4000000\left(200+t\right)^{-3}\right)\\\\1-0\\\\1$

The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$

4 0

3 years ago

An iron tank is constructed in the form of an isosceles trapezoid. Each base angle measures 105°, the length of the base is 10 f

Nikolay [14]

The right answer for the question that is being asked and shown above is that: "13 ft."An iron tank is constructed in the form of an isosceles trapezoid. Each base angle measures 105°, the length of the base is 10 feet, and each of the nonparallel sides is 12 feet long. The length of a diagonal brace to the nearest whole number is 13 ft

4 0

3 years ago

Read 2 more answers

If you flip a fair coin 5 times, and see 3 heads and 2 tails, what fraction of the time will you expect to see a head on the six

Margaret [11]

I hope I'm not wrong here, but B.1/2? Only because its a 50% chance

6 0

3 years ago