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Sergeu [11.5K]
2 years ago
9

Marco mixed 2 teaspoons of sugar and 1 teaspoon of milk into his coffee. What is the ratio of teaspoons of sugar

Mathematics
2 answers:
AlekseyPX2 years ago
6 0

Answer:

A..................... is answer

Andre45 [30]2 years ago
5 0

The correct answer is: A. 2 : 1

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(x^2y+e^x)dx-x^2dy=0
klio [65]

It looks like the differential equation is

\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0

Check for exactness:

\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0

*is* exact. If this modified DE is exact, then

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}

We have

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
Exactly how many days is 72 hours?
jasenka [17]
72 hours is exactly 3 days...
5 0
3 years ago
Read 2 more answers
In a group of 31 students, 7 study both Art and Biology.
ale4655 [162]

Answer:

it is 358

Step-by-step explanation:

Hope this helped have an amazing day!

7 0
3 years ago
50 POINTS PLS HELP!!!
Viefleur [7K]

Answer:

1.) 90 ways

2.) 15,600 ways

3.) 24 ways

Step-by-step explanation:

1.)  Let's say   we are given 2 digits not equal to each other...The number of different orderings would be  2*1 = 2, or  if we wanted the number of ways to order 2 digits chosen :   10*9 = 90 ways

2.)   The number of ways to arrange 3 letters from 26 letters is:

26 Permutes 3  =    26! / (26 -3)!  =  26*25*24 = 15,600 ways to choose 3 letters in a certain order

3.) 4 different chores.  The number of ways to do them =  4*3*2 *1 = 24 ways

7 0
3 years ago
Karla invests $300 compounded every 6 months at a rate of 10% for 3 years. At the end of three years, Karla will have $402.03 in
adell [148]

Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1 + r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $300

r = 10% = 10/100 = 0.1

n = 2 because it was compounded 2 times in a year(6 months).

t = 3 years

Therefore,

A = 300(1 + 0.1/2)^2 × 3

A = 300(1 + 0.05)^6

A = 300(1.05)^6

A = $402.03

4 0
3 years ago
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