Ask question

Ask question

All categories

3 years ago

6

90% of people marry there 7th grade love. since u have read this, u will be told good news 2night. if u don't pass this on nine

comments ur worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of ur life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love you...
Guys I believe this. Don't think it is fake. This might be the fate you have chosen...

Advanced Placement (AP)

2 answers:

emmasim [6.3K]3 years ago

4 0

Just chill........... good luck

Ira Lisetskai [31]3 years ago

3 0

Answer:

um okkkk

Explanation:

You might be interested in

Which laws protect people against criminal liability while they attempt to help another person during a crisis?

dimulka [17.4K]

The one law I am familiar with is The Good Samaritan Law.
(GSL)
For example, If one comes upon another person needing medical or first aid assistance & provides help within his knowledge then he is protected under the GSL.

5 0

4 years ago

Assume gasoline is sold in a competitive market, the equilibrium price is $50 per barrel, and the equilibrium quantity is 1000 b

sp2606 [1]

Answer:

(a) Please find attached, the required graph showing the equilibrium price at $50 per barrel

(b) There will be a shortage

(c) Please find attached, the graph showing change in demand due to increased supply from the new oil well

(d) When there is an increase in the price of gasoline operated automobiles, there would be a downward shift in demand while supply would remain constant, and the equilibrium price will decrease

(e) The combined effect will be a further reduction in the price per barrel

Explanation:

The equilibrium price

(a) Please find attached, the required graph

(b) As shown on the graph, in a competitive market, at a price of $40 per barrel, there will be a shortage in the market due to the amount of money available will be more than the total price of the number of barrel for sale, because the equilibrium price at which the amount of money available in the market is equal to the total price of the barrels supplied gives the amount of money people a willing and able to spend on a barrel

(c) If new oil is discovered, the number of barrels supplied will be more than the amount of money available to purchase a barrel and the equilibrium price will drop

(d) When there is an increase in the price of gasoline operated automobiles, there would be a downward shift in demand while supply would remain constant, and the equilibrium price will decrease

(e) If both oil is discovered and there is an increase in the price of as operated vehicles, the result will be a further reduction in the price of a barrel of oil

8 0

3 years ago

Solve the following differential equation with initial conditions: y''=e^-2t+10e^4t ; y(0)=1, y'(0)=0

skad [1K]

Answer:

Option A. $y = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

Explanation:

This is a second order DE, so we'll need to integrate twice, applying initial conditions as we go. At a couple points, we'll need to apply u-substitution.

<u>Round 1:</u>

To solve the differential equation, write it as differentials, move the differential, and integrate both sides:

$y''=e^{-2t}+10e^{4t}$

$\frac{dy'}{dt}=e^{-2t}+10e^{4t}$

$dy'=[e^{-2t}+10e^{4t}]dt$

$\int dy'=\int [e^{-2t}+10e^{4t}]dt$

Applying various properties of integration:

$\int dy'=\int e^{-2t} dt + \int 10e^{4t}dt\\\int dy'=\int e^{-2t} dt + 10\int e^{4t}dt$

Prepare for integration by u-substitution

$\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt$ , letting $u_1=-2t$ and $u_2=4t$

Find dt in terms of $u_1 \text{ and } u_2$

$u_1=-2t\\du_1=-2dt\\-\frac{1}{2}du_1=dt$ $u_2=4t\\du_2=4dt\\\frac{1}{4}du_2=dt$

$\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt\\\int dy'=\int e^{u_1} (-\frac{1}{2} du_1) + 10\int e^{u_2} (\frac{1}{4} du_2)\\\int dy'=-\frac{1}{2} \int e^{u_1} (du_1) + 10 *\frac{1}{4} \int e^{u_2} (du_2)$

Using the Exponential rule (don't forget your constant of integration):

$y'=-\frac{1}{2} e^{u_1} + 10 *\frac{1}{4}e^{u_2} +C_1$

Back substituting for $u_1 \text{ and } u_2$ :

$y'=-\frac{1}{2} e^{(-2t)} + 10 *\frac{1}{4}e^{(4t)} +C_1\\y'=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\$

<u>Finding the constant of integration</u>

Given initial condition $y'(0)=0$

$y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\0=y'(0)=-\frac{1}{2} e^{-2(0)} + \frac{5}{2}e^{4(0)} +C_1\\0=-\frac{1}{2} (1) + \frac{5}{2}(1) +C_1\\-2=C_1\\$

The first derivative with the initial condition applied: $y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\$

<u>Round 2:</u>

Integrate again:

$y' =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\\frac{dy}{dt} =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\dy =[-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int [-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int -\frac{1}{2} e^{-2t} dt + \int \frac{5}{2}e^{4t} dt - \int 2 dt\\\int dy = -\frac{1}{2} \int e^{-2t} dt + \frac{5}{2} \int e^{4t} dt - 2 \int dt\\$

$y = -\frac{1}{2} * -\frac{1}{2} e^{-2t} + \frac{5}{2} * \frac{1}{4} e^{4t} - 2 t + C_2\\y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + C_2$

<u />

<u>Finding the constant of integration :</u>

Given initial condition $y(0)=1$

$1=y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + C_2\\1 = \frac{1}{4} (1) + \frac{5}{8} (1) - (0) + C_2\\1 = \frac{7}{8} + C_2\\\frac{1}{8}=C_2$

So, $y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

<u>Checking the solution</u>

$y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

This matches our initial conditions here $y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + \frac{1}{8} = 1$

Going back to the function, differentiate:

$y' = [\frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}]'\\y' = [\frac{1}{4} e^{-2t}]' + [\frac{5}{8} e^{4t}]' - [2 t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} [e^{-2t}]' + \frac{5}{8} [e^{4t}]' - 2 [t]' + [\frac{1}{8}]'$

Apply Exponential rule and chain rule, then power rule

$y' = \frac{1}{4} e^{-2t}[-2t]' + \frac{5}{8} e^{4t}[4t]' - 2 [t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} e^{-2t}(-2) + \frac{5}{8} e^{4t}(4) - 2 (1) + (0)\\y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2$

This matches our first order step and the initial conditions there.

$y'(0) = -\frac{1}{2} e^{-2(0)} + \frac{5}{2} e^{4(0)} - 2=0$

Going back to the function y', differentiate:

$y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2\\y'' = [-\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2]'\\y'' = [-\frac{1}{2} e^{-2t}]' + [\frac{5}{2} e^{4t}]' - [2]'\\y'' = -\frac{1}{2} [e^{-2t}]' + \frac{5}{2} [e^{4t}]' - [2]'$

Applying the Exponential rule and chain rule, then power rule

$y'' = -\frac{1}{2} e^{-2t}[-2t]' + \frac{5}{2} e^{4t}[4t]' - [2]'\\y'' = -\frac{1}{2} e^{-2t}(-2) + \frac{5}{2} e^{4t}(4) - (0)\\y'' = e^{-2t} + 10 e^{4t}$

So our proposed solution is a solution to the differential equation, and satisfies the initial conditions given.

7 0

2 years ago

If the x axis is quantity, and the y axis is price, what does line s1 represent

diamong [38]

The increase in price as the quality increase

7 0

4 years ago

Which table shows a function that is decreasing only over the interval (–1, ∞)?

Natasha2012 [34]

Answer: The table to the farthest right.

Explanation:

The interval (-1, ∞) represents x-values anywhere from -1 and higher.

As such, you are looking for a table with y-values that decrease as the x-values increase.

Starting from -1 -

The first option: -5, -2, -1, 2. This is increasing.

The second option: -7, -6, 1, -1. This increases from -7 to -6 to 1.

The third option: -1, 2, 1, -6. This increases from -1 to 2.

The fourth option: 1, 0, -4, -8. This decreases across all coordinates in the range specified.

5 0

3 years ago

Read 2 more answers