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2 years ago

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During the 2008-2009 financial crisis, the GDP of the USA reduced from $14.72 trillion in 2008 to$14.42 trillion in 2009. What i

s the percentage change between these two years? Round your answer to the nearest hundredth of a percent. Do not include the percent sign in your answer. For example, if your answer is 15.57%, enter 15.57.

1 answer:

Bad White [126]2 years ago

4 0

Answer:

2.04

Step-by-step explanation:

The GDP of USA IN 2008 = $14.72 trillion

The GDP of USA IN 2009 = $14.42 trillion

Change in GDP

$14.72 trillion- $14.42 trillion = $0.3 trillion

Percentage Change

($0.3 trillion/$14.72 trillion) x 100%

0.02038 X 100% = 2.038=2.04 approximately

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The greatest amount of money that could be rounded to $105.40 would be $105.44. Any higher, and it would be rounded to $105.50.

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Rebecca went swimming yesterday. After a while she had covered one fifth of her intended distance. After swimming six more lengt

alukav5142 [94]

Answer:

Option E.

Step-by-step explanation:

Let the intended length of Rebecca's swimming is = x units

and we assume the length of the pool = l units

Now it is given in the question that " She covers one fifth of her intended distance "

That means distance covered = $\frac{x}{5}$

" After swimming six more lengths of the pool she had covered one quarter of her intended distance"

So $\frac{x}{5}+6(l)=\frac{x}{4}$

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x = 20×(6l)

x = 120l

Therefore, Rebecca has to complete 120 lengths of the pool.

Option E is the answer.

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2 years ago

Two solutions of different concentrations of acid are mixed creating 40 mL of a solution that is 32% acid. One-quarter of the so

Georgia [21]

In order to construct this equation, we will use the variables:
V to represent mixture volume (40 ml)
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v₂ to represent the volume of the second solution (40 * 3/4 = 30 ml)
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We know that the total amount of substance, product of the volume and concentration, in the final solution is equal to the individual amounts in the two given solutions. Thus:
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A 400 gallon tank initially contains 100 gal of brine containing 50 pounds of salt. Brine containing 1 pound of salt per gallon

posledela

Answer:

The amount of salt in the tank when it is full of brine is 393.75 pounds.

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 <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> 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 y<em>(t)</em> be the amount of salt (in pounds) in the tank at time <em>t</em> (in seconds). Then we can represent the situation with the below picture.

Then the differential equation we’re after is

$\frac{dy}{dt} = (Rate \:in)- (Rate \:out)\\\\\frac{dy}{dt} = 5 \:\frac{gal}{s} \cdot 1 \:\frac{pound}{gal}-3 \:\frac{gal}{s}\cdot \frac{y(t)}{V(t)} \:\frac{pound}{gal}\\\\\frac{dy}{dt} =5\:\frac{pound}{s}-3 \frac{y(t)}{V(t)} \:\frac{pound}{s}$

$V(t)$ is the volume of brine in the tank at time <em>t. </em>To find it we know that at time 0 there were 100 gallons, 5 gallons are added and 3 are drained, and the net increase is 2 gallons per second. So,

$V(t)=100 + 2t$

We can then write the initial value problem:

$\frac{dy}{dt} =5-\frac{3y}{100+2t} , \quad y(0)=50$

We have a linear differential equation. A first-order linear differential equation is one that can be put into the form

$\frac{dy}{dx}+P(x)y =Q(x)$

where <em>P</em> and <em>Q</em> are continuous functions on a given interval.

In our case, we have that

$\frac{dy}{dt}+\frac{3y}{100+2t} =5 , \quad y(0)=50$

The solution process for a first order linear differential equation is as follows.

Step 1: Find the integrating factor, $\mu \left( x \right)$ , using $\mu \left( x \right) = \,{{\bf{e}}^{\int{{P\left( x \right)\,dx}}}$

$\mu \left( t \right) = \,{{e}}^{\int{{\frac{3}{100+2t}\,dt}}}\\\int \frac{3}{100+2t}dt=\frac{3}{2}\ln \left|100+2t\right|\\\\\mu \left( t \right) =e^{\frac{3}{2}\ln \left|100+2t\right|}\\\\\mu \left( t \right) =(100+2t)^{\frac{3}{2}$

Step 2: Multiply everything in the differential equation by $\mu \left( x \right)$ and verify that the left side becomes the product rule $\left( {\mu \left( t \right)y\left( t \right)} \right)'$ and write it as such.

$\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+\frac{3y}{100+2t}\cdot \left(100+2t\right)^{\frac{3}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}\cdot \left(100+2t\right)^{\frac{3}{2}}+3y\cdot \left(100+2t\right)^{\frac{1}{2}}=5 \left(100+2t\right)^{\frac{3}{2}}\\\\\frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})=5\left(100+2t\right)^{\frac{3}{2}}$

Step 3: Integrate both sides.

$\int \frac{dy}{dt}(y \left(100+2t\right)^{\frac{3}{2}})dt=\int 5\left(100+2t\right)^{\frac{3}{2}}dt\\\\y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }+ C$

Step 4: Find the value of the constant and solve for the solution y(t).

$50 \left(100+2(0)\right)^{\frac{3}{2}}=(100+2(0))^{\frac{5}{2} }+ C\\\\100000+C=50000\\\\C=-50000$

$y \left(100+2t\right)^{\frac{3}{2}}=(100+2t)^{\frac{5}{2} }-50000\\\\y(t)=100+2t-\frac{50000}{\left(100+2t\right)^{\frac{3}{2}}}$

Now, the tank is full of brine when:

$V(t) = 400\\100+2t=400\\t=150$

The amount of salt in the tank when it is full of brine is

$y(150)=100+2(150)-\frac{50000}{\left(100+2(150)\right)^{\frac{3}{2}}}\\\\y(150)=393.75$

6 0

2 years ago

Hans spent three days baking 224 cookies. On the second day, he baked twice as many as he did on the first day. On the third day

sergeinik [125]

Answer:

222 im my answer

Step-by-step explanation:

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