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4 years ago

5

Can someone help me asap please

1 answer:

lutik1710 [3]4 years ago

8 0

Answer:

The answer is $19.44

Step-by-step explanation:

First you have to find 25% of $28.80 (which is 7.20). Then, you subtract the $7.20 from your 28.80. You do that because you are getting a 25% discount. Then, you find 10% of $21.60 (your answer for the 25% discount) and you subtract 10% of $21.60. After that, you get your total of $19.44.

I hope you find this helpful.

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tatuchka [14]

Answer:

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Step-by-step explanation:

The point that they meet is 2,4 . It means X+ is 2 and Y+ is 4 .

You can write a as their meet point . you should answer like this a(2,4) .

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3 years ago

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4 years ago

Read 2 more answers

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

so

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

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3 years ago

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chubhunter [2.5K]

Answer:

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Step-by-step explanation:

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Multiply it by 5

= 2.738612788

Square it

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Hope this helped. :)

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3 years ago

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Alja [10]

Answer:

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Step-by-step explanation:

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3 years ago