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3 years ago

If 5 hamburgers cost $10, what is the cost for 3 hamburgers?

Mathematics

1 answer:

mestny [16]3 years ago

6 0

Answer: 6$

Step-by-step explanation: 5 divided by 10 = 2$ so 3 x 2 = 6

I do not hold responsibility if you get it wrong :(

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What is the answer to this problem? <br><br> 2 1/3 + 3 1/3 =?

Naily [24]

Answer:

Convert the mixed numbers to improper fractions, then find the LCD and combine.

Exact Form:

173

Decimal Form:

5.¯6

Mixed Number Form:

523

Step-by-step explanation:

3 0

3 years ago

David's optometrist sold him a bottle of eyeglass cleaner containing 30 ml of glasscleaning solution. assuming there are 20 drop

Furkat [3]

Each complete cleaning requires 6 drops of cleaner. Let's convert that to ml:

6 drops 1 ml 3 ml cleaner
----------- * ------------- = ------------------- = 0.3 ml cleaner per cleaning
20 drops 10 cleanings

Recall that 1 bottle contains 30 ml cleaning fluid.

How many cleanings is one bottle of fluid good for?

1 bottle 0.3 ml
----------- = ----------- => x = 100 cleanings/bottle
x 30 ml

8 0

3 years ago

At a potato chip factory there were 92 machines working with each machine able to produce 56 chips a minute.If this is enough po

Mademuasel [1]

Divide 56 by 2 and you get 28,therefore 28 chips per box

3 0

3 years ago

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.