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2 years ago

An energy _________________ occurs when energy changes from one type to another type.

Mathematics

1 answer:

ZanzabumX [31]2 years ago

8 0

Answer:

transformation

Step-by-step explanation:

this is when one form of energy forms into another form, and it can be transferred into a new object or location

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PLEASE HELP QUICKLY!!! A student's parents offer him one of two options to receive scholarship money for college.

emmainna [20.7K]

Answer:

A = $47,500.00, B = $65,536.00; B, because he receives $18,036.00 more

Step-by-step explanation:

Option A Versus Option B

Option A Option B

Day Amount Deposited Day Amount Deposited

0 $40,000.00 0 $1.00

1 $40,250.00 1 $2.00

2 $40,500.00 2 $4.00

...

30 $47,500 16 $65,536.00

Answer:

A = $47,500.00, B = $65,536.00; B, because he receives $18,036.00 more

7 0

2 years ago

A company sells widgets. The amount of profit, y, made by the company, is related to the selling price of each widget, x, by the

Alika [10]

Your answer is 5173.

6 0

2 years ago

F (x) = 3x^3 - 4x^2 + 6x and g (x) = 5x^3 + 2x^2 - 3x. What is f (x) - g (x)

nekit [7.7K]

9x2-1............................................................................

5 0

2 years ago

Shawna won 40 super bouncy balls playing

patriot [66]

Answer:

13 not including Shawna

Step-by-step explanation:

First, you have to take into account that the remainder is 1, so you must subtract 1 from 40, giving you 39. 39/3 is 13, so 13 is your final answer.

5 0

1 year ago

Read 2 more answers

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

2 years ago